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Concept Review

Differential Calculus

Extreme Values and the Mean Value Theorem

Optimization is the payoff of differential calculus: derivatives exist largely so that we can find the highest and lowest values a quantity can take. This topic supplies the vocabulary (absolute vs. local extrema, critical numbers), the existence theorems that guarantee you are not hunting for something that is not there (the Extreme Value Theorem, Rolle, the Mean Value Theorem), and the finite checklist — the Closed Interval Method — that turns the search into an algorithm. The Mean Value Theorem then closes the loop back to theory: nearly every later statement about ff deduced from f′f' is a corollary of it.

Absolute vs. local extrema

Let ff have domain DD and let cc be a number in DD.

  • ff has an absolute (global) maximum at cc if f(c)≥f(x)f(c)\ge f(x) for every xx in DD. It has an absolute minimum at cc if f(c)≤f(x)f(c)\le f(x) for every xx in DD.
  • ff has a local (relative) maximum at cc if f(c)≥f(x)f(c)\ge f(x) for all xx in some open interval that contains cc and lies inside DD. Local minimum: reverse the inequality.
  • The extreme value is the number f(c)f(c); the location is cc. A complete answer names both: "the absolute maximum value is 55, attained at x=−1x=-1."

Two consequences of the definitions. An absolute extremum that occurs at an interior point of DD is automatically a local extremum. With the "open interval" convention above, an endpoint of a closed-interval domain is never a local extremum (no open interval around it fits inside DD) — but it may perfectly well be an absolute one, which is exactly why the Closed Interval Method tests endpoints separately.

Critical numbers

Definition. A number cc is a critical number of ff if cc is an interior point of the domain of ff and either

f′(c)=0orf′(c) does not exist.f'(c)=0 \qquad\text{or}\qquad f'(c) \text{ does not exist.}

Both branches are compulsory. Students lose marks almost exclusively on the second one.

Functionf′(x)f'(x)Critical numbersWhat happens there
x3−3xx^{3}-3x3x2−33x^{2}-3x=−1, 1x=-1,\ 1local max, local min
x2/3x^{2/3}23x−1/3\dfrac{2}{3}x^{-1/3}x=0x=0cusp, absolute min
x1/3x^{1/3}13x−2/3\dfrac{1}{3}x^{-2/3}x=0x=0vertical tangent, no extremum
x3x^{3}3x23x^{2}x=0x=0stationary inflection, no extremum
1/x1/x−1/x2-1/x^{2}none00 is not in the domain, so it is not critical

Fermat's Theorem

Theorem (Fermat). If ff has a local maximum or a local minimum at cc, and cc is an interior point of the domain, and f′(c)f'(c) exists, then f′(c)=0f'(c)=0.

Why. Suppose a local max at cc. For small h>0h>0, f(c+h)−f(c)≤0f(c+h)-f(c)\le 0, so f(c+h)−f(c)h≤0\dfrac{f(c+h)-f(c)}{h}\le 0 and the right-hand limit is ≤0\le 0. For small h<0h<0 the same numerator is ≤0\le 0 but h<0h<0, so the quotient is ≥0\ge 0 and the left-hand limit is ≥0\ge 0. Because f′(c)f'(c) exists, the two one-sided limits both equal f′(c)f'(c), forcing f′(c)≤0f'(c)\le 0 and f′(c)≥0f'(c)\ge 0, hence f′(c)=0f'(c)=0.

The converse is false. f′(c)=0f'(c)=0 does not imply an extremum at cc: for f(x)=x3f(x)=x^{3}, f′(0)=0f'(0)=0, yet f′(x)=3x2>0f'(x)=3x^{2}>0 for every x≠0x\ne 0, so ff is strictly increasing and 00 is not an extremum. Nor does an extremum force f′(c)=0f'(c)=0 — it may not exist, as at x=0x=0 for f(x)=∣x∣f(x)=|x|.

What Fermat is actually used for is its contrapositive: an extremum can only occur at a critical number or at an endpoint of the domain. That is a search strategy, not a test.

The Extreme Value Theorem (EVT)

Theorem. If ff is continuous on a closed and bounded interval [a,b][a,b], then ff attains an absolute maximum value f(c)f(c) and an absolute minimum value f(d)f(d) for some numbers cc and dd in [a,b][a,b].

Every hypothesis is load-bearing:

Hypothesis droppedExampleFailure
closedf(x)=xf(x)=x on (0,1)(0,1)neither max nor min attained
boundedf(x)=xf(x)=x on [0,∞)[0,\infty)no absolute maximum
continuousf(x)=x2f(x)=x^{2} for 0≤x<20\le x<2 together with f(2)=0f(2)=0, on [0,2][0,2]values climb toward 44 but never reach it: no maximum

EVT is sufficient, not necessary: a discontinuous function can still have both extrema. It is also purely an existence theorem — it never tells you where.

The Closed Interval Method

For ff continuous on [a,b][a,b]:

  1. Find all critical numbers of ff that lie in the open interval (a,b)(a,b).
  2. Evaluate ff at each of them.
  3. Evaluate f(a)f(a) and f(b)f(b).
  4. The largest of these values is the absolute maximum; the smallest is the absolute minimum.

It is valid because EVT says the extrema exist, and Fermat says an interior extremum must sit at a critical number — so the finite list is exhaustive. It fails if ff is not continuous on all of [a,b][a,b], or if ff has infinitely many critical numbers there.

Worked Example 1 — both types of critical number

Find the absolute extrema of f(x)=3x2/3−2xf(x)=3x^{2/3}-2x on [−1,8][-1,8].

ff is continuous on [−1,8][-1,8] because x2/3=(x1/3)2x^{2/3}=\left(x^{1/3}\right)^{2} is defined and continuous for all real xx. Differentiate:

f′(x)=3⋅23x−1/3−2=2x1/3−2=2(1−x1/3)x1/3.\begin{aligned} f'(x) &= 3\cdot\frac{2}{3}x^{-1/3}-2 \\ &= \frac{2}{x^{1/3}}-2 \\ &= \frac{2\left(1-x^{1/3}\right)}{x^{1/3}} . \end{aligned}

f′(x)=0f'(x)=0 requires x1/3=1x^{1/3}=1, i.e. x=1x=1. And f′(x)f'(x) fails to exist at x=0x=0, which is in the domain — so x=0x=0 is also a critical number (a cusp). Critical numbers: x=0x=0 and x=1x=1.

f(−1)=3(−1)2/3−2(−1)=3(1)+2=5f(0)=0f(1)=3(1)−2(1)=1f(8)=3(4)−2(8)=12−16=−4.\begin{aligned} f(-1) &= 3(-1)^{2/3}-2(-1)=3(1)+2=5 \\ f(0) &= 0 \\ f(1) &= 3(1)-2(1)=1 \\ f(8) &= 3(4)-2(8)=12-16=-4 . \end{aligned}

Absolute maximum 55 at x=−1x=-1; absolute minimum −4-4 at x=8x=8. Note that both occur at endpoints and neither at a critical number — losing the endpoints would have produced two wrong answers. (For the record, the sign of f′f' shows x=0x=0 is a local minimum and x=1x=1 a local maximum.)

Rolle's Theorem

Theorem (Rolle). If ff is continuous on [a,b][a,b], differentiable on (a,b)(a,b), and f(a)=f(b)f(a)=f(b), then there is at least one cc with a<c<ba<c<b such that f′(c)=0f'(c)=0.

Why. By EVT, ff has an absolute max and min on [a,b][a,b]. If both occur at endpoints then, since f(a)=f(b)f(a)=f(b), the max and min values are equal, so ff is constant and f′(c)=0f'(c)=0 for every interior cc. Otherwise some extremum occurs at an interior point, and Fermat gives f′(c)=0f'(c)=0 there.

Drop differentiability and it dies: f(x)=1−x2/3f(x)=1-x^{2/3} on [−1,1][-1,1] has f(−1)=f(1)=0f(-1)=f(1)=0 and is continuous, but f′(x)=−23x−1/3f'(x)=-\frac{2}{3}x^{-1/3} is never 00 and does not exist at x=0x=0.

The Mean Value Theorem (MVT)

Theorem (MVT). If ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b), then there is at least one cc with a<c<ba<c<b such that

f′(c)=f(b)−f(a)b−a,equivalentlyf(b)−f(a)=f′(c)(b−a).f'(c)=\frac{f(b)-f(a)}{b-a}, \qquad\text{equivalently}\qquad f(b)-f(a)=f'(c)(b-a).

Geometrically: somewhere strictly inside the interval, the tangent line is parallel to the secant line through (a,f(a))\left(a,f(a)\right) and (b,f(b))\left(b,f(b)\right). Physically: at some instant your instantaneous speed equals your average speed over the trip.

Proof. Let L(x)=f(a)+f(b)−f(a)b−a(x−a)L(x)=f(a)+\dfrac{f(b)-f(a)}{b-a}(x-a) be the secant line and set h(x)=f(x)−L(x)h(x)=f(x)-L(x). Then hh inherits continuity on [a,b][a,b] and differentiability on (a,b)(a,b), and h(a)=h(b)=0h(a)=h(b)=0. Rolle gives cc with h′(c)=0h'(c)=0, i.e. f′(c)=f(b)−f(a)b−af'(c)=\dfrac{f(b)-f(a)}{b-a}.

Rolle is the special case f(a)=f(b)f(a)=f(b). MVT asserts existence only: cc need not be unique and usually cannot be found in closed form.

Worked Example 2 — finding cc, and using MVT as a tool

(a) For f(x)=x3−xf(x)=x^{3}-x on [0,2][0,2]: ff is a polynomial, hence continuous on [0,2][0,2] and differentiable on (0,2)(0,2). The secant slope is

f(2)−f(0)2−0=(8−2)−(0)2=62=3.\frac{f(2)-f(0)}{2-0}=\frac{(8-2)-(0)}{2}=\frac{6}{2}=3 .

Solve f′(c)=3c2−1=3f'(c)=3c^{2}-1=3: c2=43c^{2}=\dfrac{4}{3}, so c=±23=±233c=\pm\dfrac{2}{\sqrt3}=\pm\dfrac{2\sqrt3}{3}. Only c=233≈1.1547c=\dfrac{2\sqrt3}{3}\approx 1.1547 lies in (0,2)(0,2). Check: 3(43)−1=4−1=33\left(\frac43\right)-1=4-1=3. ✔

(b) Prove ∣sin⁡b−sin⁡a∣≤∣b−a∣|\sin b-\sin a|\le|b-a| for all real a,ba,b. If a=ba=b it is trivial; otherwise apply MVT to f(x)=sin⁡xf(x)=\sin x (continuous and differentiable everywhere) on the interval with endpoints aa and bb: sin⁡b−sin⁡a=(cos⁡c)(b−a)\sin b-\sin a=(\cos c)(b-a) for some cc between them. Since ∣cos⁡c∣≤1|\cos c|\le 1,

∣sin⁡b−sin⁡a∣=∣cos⁡c∣⋅∣b−a∣≤∣b−a∣.|\sin b-\sin a|=|\cos c|\cdot|b-a|\le|b-a| .

Consequences of the MVT

  1. Zero derivative ⇒\Rightarrow constant. If f′(x)=0f'(x)=0 for all xx in an interval II, then ff is constant on II. Proof: pick x1<x2x_1<x_2 in II; MVT on [x1,x2][x_1,x_2] gives f(x2)−f(x1)=f′(c)(x2−x1)=0f(x_2)-f(x_1)=f'(c)(x_2-x_1)=0, so f(x2)=f(x1)f(x_2)=f(x_1).
  2. Equal derivatives ⇒\Rightarrow differ by a constant. If f′(x)=g′(x)f'(x)=g'(x) for all xx in an interval II, apply (1) to f−gf-g: f(x)=g(x)+Cf(x)=g(x)+C on II. This is precisely why the antiderivative carries "+C+C".
  3. Derivative bounds transfer to the function. If ff is continuous on [a,b][a,b], differentiable on (a,b)(a,b), and m≤f′(x)≤Mm\le f'(x)\le M for every xx in (a,b)(a,b), then m(b−a)≤f(b)−f(a)≤M(b−a)m(b-a)\le f(b)-f(a)\le M(b-a). Proof: MVT gives f(b)−f(a)=f′(c)(b−a)f(b)-f(a)=f'(c)(b-a) for some cc in (a,b)(a,b), and b−a>0b-a>0, so the bounds on f′(c)f'(c) carry across. The continuity hypothesis is not decoration: without it the endpoint values f(a)f(a) and f(b)f(b) are not controlled by f′f' at all.

The word "interval" in (1) and (2) is a hypothesis, not decoration. On the domain of f(x)=∣x∣xf(x)=\dfrac{|x|}{x} — the union of (−∞,0)(-\infty,0) and (0,∞)(0,\infty), which is not an interval — f′(x)=0f'(x)=0 everywhere, yet ff takes the values −1-1 and 11. All you may conclude is that ff is constant on each separate interval.

Common mistakes

WrongRight
"Critical numbers are where f′(x)=0f'(x)=0."Also where f′(c)f'(c) fails to exist, provided cc is an interior point of the domain: f(x)=x2/3f(x)=x^{2/3} has critical number 00.
For f(x)=xx−2f(x)=\dfrac{x}{x-2}, calling x=2x=2 a critical number because f′f' blows up there.22 is not in the domain, so it is not a critical number.
Using the Closed Interval Method but forgetting f(a)f(a) and f(b)f(b).Endpoints often are the answer — see Worked Example 1.
"f′(c)=0f'(c)=0, therefore ff has a local extremum at cc."That is the false converse of Fermat: x3x^{3} at c=0c=0. You need a First/Second Derivative Test.
Applying Rolle to f(x)=x2/3f(x)=x^{2/3} on [−1,1][-1,1] and concluding some f′(c)=0f'(c)=0.ff is not differentiable at 00, so the hypothesis fails; no such cc exists — the graph has a cusp at 00, not a horizontal tangent.
Applying MVT to f(x)=1xf(x)=\dfrac1x on [−1,1][-1,1].ff is not continuous on [−1,1][-1,1] (it is not even defined at 00); the "conclusion" f′(c)=1f'(c)=1 (the secant slope) has no solution, since f′(x)=−1/x2<0f'(x)=-1/x^{2}<0 always.
Reporting the location as the extreme value.Max value is f(c)f(c); the max occurs at cc. State both.
Assuming MVT's cc is unique, or solvable.MVT is an existence theorem; x3−xx^{3}-x on [−2,2][-2,2] has two valid cc's.
"f′=g′f'=g' everywhere on their domains, so f=g+Cf=g+C."Only on an interval. arctan⁡x\arctan x and −arctan⁡1x-\arctan\frac1x have equal derivatives off 00 but differ by π2\frac{\pi}{2} on one side and −π2-\frac{\pi}{2} on the other.

Key terms

  • absolute (global) maximum
  • absolute (global) minimum
  • local (relative) extremum
  • critical number
  • stationary point
  • cusp / vertical tangent
  • Fermat's Theorem
  • false converse of Fermat
  • Extreme Value Theorem
  • closed and bounded interval
  • Closed Interval Method
  • endpoint evaluation
  • Rolle's Theorem
  • Mean Value Theorem
  • secant line vs. tangent line
  • average rate of change
  • zero derivative implies constant
  • functions with equal derivatives
  • existence theorem

Practice Problems

Find all critical numbers of f(x)=2x3−3x2−12x+5f(x)=2x^{3}-3x^{2}-12x+5.

Show hint

Every critical number is either a solution of f′(x)=0f'(x)=0 or a point of the domain where f′f' fails to exist. This ff is a polynomial, so only one of those two possibilities can occur.

Show answer

A polynomial is differentiable at every real number, so f′f' never fails to exist: the only critical numbers come from f′(x)=0f'(x)=0.

f′(x)=6x2−6x−12.f'(x)=6x^{2}-6x-12 .

Factor:

6x2−6x−12=6(x2−x−2)=6(x−2)(x+1).\begin{aligned} 6x^{2}-6x-12 &= 6\left(x^{2}-x-2\right) \\ &= 6(x-2)(x+1) . \end{aligned}

Setting this to 00 gives x=2x=2 and x=−1x=-1, both in the domain (all real numbers).

Check by substitution: f′(−1)=6−(−6)−12=6+6−12=0f'(-1)=6-(-6)-12=6+6-12=0 ✔ and f′(2)=24−12−12=0f'(2)=24-12-12=0 ✔.

Critical numbers: x=−1x=-1 and x=2x=2.

(For orientation, the sign of f′=6(x−2)(x+1)f'=6(x-2)(x+1) is +,−,++,-,+ on (−∞,−1)(-\infty,-1), (−1,2)(-1,2), (2,∞)(2,\infty), so ff has a local maximum f(−1)=12f(-1)=12 and a local minimum f(2)=−15f(2)=-15.)

Find all critical numbers of g(x)=x4/3−4x1/3g(x)=x^{4/3}-4x^{1/3}, and state which type each one is (g′=0g'=0 or g′g' undefined).

Show hint

Differentiate with the power rule, then factor out the most negative power of xx so you can see both where the derivative vanishes and where it blows up.

Show answer

The domain of gg is all real numbers, since x1/3x^{1/3} is defined for every real xx and x4/3=(x1/3)4x^{4/3}=\left(x^{1/3}\right)^{4}.

g′(x)=43x1/3−4⋅13x−2/3=43x1/3−43x−2/3=43x−2/3(x1/3+2/3−1)=4(x−1)3x2/3.\begin{aligned} g'(x) &= \frac{4}{3}x^{1/3}-4\cdot\frac{1}{3}x^{-2/3} \\ &= \frac{4}{3}x^{1/3}-\frac{4}{3}x^{-2/3} \\ &= \frac{4}{3}x^{-2/3}\left(x^{1/3+2/3}-1\right) \\ &= \frac{4(x-1)}{3x^{2/3}} . \end{aligned}

Where g′(x)=0g'(x)=0: the numerator vanishes, x=1x=1.

Where g′(x)g'(x) does not exist: the denominator 3x2/33x^{2/3} vanishes at x=0x=0. Since x=0x=0 is in the domain of gg, it is a critical number too.

Critical numbers: x=0x=0 (derivative undefined) and x=1x=1 (derivative zero).

Verification of the factoring at a sample point, say x=8x=8: from the factored form g′(8)=4(7)3⋅4=2812=73g'(8)=\dfrac{4(7)}{3\cdot 4}=\dfrac{28}{12}=\dfrac73; from the unfactored form 43(2)−43(14)=83−13=73\dfrac43(2)-\dfrac43\left(\dfrac14\right)=\dfrac83-\dfrac13=\dfrac73 ✔.

Worth noticing: 3x2/3>03x^{2/3}>0 for all x≠0x\ne 0, so g′g' has the sign of x−1x-1. Thus gg decreases on (−∞,1)(-\infty,1) and increases on (1,∞)(1,\infty): x=1x=1 gives a local (indeed absolute) minimum g(1)=−3g(1)=-3, while x=0x=0 is a critical number that is not an extremum — gg simply passes through a vertical tangent while still decreasing.

Use the Closed Interval Method to find the absolute maximum and absolute minimum values of f(x)=x3−3x2+1f(x)=x^{3}-3x^{2}+1 on [−12,4]\left[-\frac12,4\right].

Show hint

First confirm the hypothesis that makes the method legal, then list the critical numbers that actually lie strictly inside the interval — and do not forget the two endpoints.

Show answer

Step 0 (hypothesis). ff is a polynomial, so it is continuous on the closed bounded interval [−12,4]\left[-\frac12,4\right]. By the Extreme Value Theorem both absolute extrema exist, and the Closed Interval Method will find them.

Step 1 (critical numbers).

f′(x)=3x2−6x=3x(x−2),f'(x)=3x^{2}-6x=3x(x-2),

so f′(x)=0f'(x)=0 at x=0x=0 and x=2x=2. Both lie in (−12,4)\left(-\frac12,4\right). There are no points where f′f' fails to exist.

Step 2 (evaluate at critical numbers and endpoints).

f(−12)=−18−3(14)+1=−18−34+1=18f(0)=1f(2)=8−12+1=−3f(4)=64−48+1=17.\begin{aligned} f\left(-\frac12\right) &= -\frac18-3\left(\frac14\right)+1=-\frac18-\frac34+1=\frac18 \\ f(0) &= 1 \\ f(2) &= 8-12+1=-3 \\ f(4) &= 64-48+1=17 . \end{aligned}

Step 3 (compare). The four values are 18, 1, −3, 17\frac18,\ 1,\ -3,\ 17.

Absolute maximum value 1717, attained at x=4x=4; absolute minimum value −3-3, attained at x=2x=2.

Independent check on the shape: f′=3x(x−2)f'=3x(x-2) is positive on (−12,0)\left(-\frac12,0\right), negative on (0,2)(0,2), positive on (2,4)(2,4). So ff rises to a local max at x=0x=0, falls to a local min at x=2x=2, then rises to the right endpoint — consistent with the minimum at x=2x=2 and the maximum at the right endpoint.

Find the absolute maximum and minimum values of f(x)=x4−x2f(x)=x\sqrt{4-x^{2}} on its natural domain [−2,2][-2,2].

Show hint

Use the product rule, then combine the two terms over the common denominator 4−x2\sqrt{4-x^{2}} before solving. Ask separately where the derivative is zero and where it blows up.

Show answer

Domain and continuity. 4−x2≥04-x^{2}\ge 0 exactly when −2≤x≤2-2\le x\le 2, so the natural domain is the closed bounded interval [−2,2][-2,2], on which ff is continuous. EVT applies.

Derivative. By the product rule with ddx4−x2=−x4−x2\dfrac{d}{dx}\sqrt{4-x^{2}}=\dfrac{-x}{\sqrt{4-x^{2}}},

f′(x)=1⋅4−x2+x⋅−x4−x2=(4−x2)−x24−x2=4−2x24−x2,−2<x<2.\begin{aligned} f'(x) &= 1\cdot\sqrt{4-x^{2}}+x\cdot\frac{-x}{\sqrt{4-x^{2}}} \\ &= \frac{\left(4-x^{2}\right)-x^{2}}{\sqrt{4-x^{2}}} \\ &= \frac{4-2x^{2}}{\sqrt{4-x^{2}}} , \qquad -2<x<2 . \end{aligned}

Critical numbers. f′(x)=0f'(x)=0 when 4−2x2=04-2x^{2}=0, i.e. x2=2x^{2}=2, i.e. x=±2x=\pm\sqrt2. Both lie in (−2,2)(-2,2). The derivative also fails to exist at x=±2x=\pm 2, but those are the endpoints of the domain, not interior points — the method evaluates them anyway.

Evaluate.

f(−2)=−20=0f(−2)=−2⋅4−2=−2⋅2=−2f(2)=2⋅2=2f(2)=20=0.\begin{aligned} f(-2) &= -2\sqrt{0}=0 \\ f\left(-\sqrt2\right) &= -\sqrt2\cdot\sqrt{4-2}=-\sqrt2\cdot\sqrt2=-2 \\ f\left(\sqrt2\right) &= \sqrt2\cdot\sqrt2=2 \\ f(2) &= 2\sqrt{0}=0 . \end{aligned}

Absolute maximum value 22 at x=2x=\sqrt2; absolute minimum value −2-2 at x=−2x=-\sqrt2.

Check: ff is odd, since f(−x)=−x4−x2=−f(x)f(-x)=-x\sqrt{4-x^{2}}=-f(x), so the extreme values must be negatives of each other and occur at opposite points — they do. Numerically, f(1.4)=1.42.04≈1.9996<2f(1.4)=1.4\sqrt{2.04}\approx 1.9996<2 ✔.

For each of the following, decide whether the Extreme Value Theorem guarantees that ff attains an absolute maximum and an absolute minimum. If it does not, name the hypothesis that fails and say whether the extrema happen to exist anyway.

(a) f(x)=1xf(x)=\dfrac1x on [1,3][1,3].

(b) f(x)=1xf(x)=\dfrac1x on (0,3](0,3].

(c) f(x)=x2f(x)=x^{2} on (−1,1)(-1,1).

(d) f(x)={x2,0≤x<20,x=2f(x)=\begin{cases} x^{2}, & 0\le x<2 \\ 0, & x=2\end{cases} on [0,2][0,2].

Show hint

EVT has exactly three hypotheses: the interval is closed, the interval is bounded, and ff is continuous on it. Check them one at a time — and remember EVT is a one-way guarantee, not a test.

Show answer

(a) f(x)=1/xf(x)=1/x on [1,3][1,3]. The only discontinuity of 1/x1/x is at x=0x=0, which is not in [1,3][1,3]; the interval is closed and bounded. All hypotheses hold, so EVT applies. Since f′(x)=−1/x2<0f'(x)=-1/x^{2}<0, ff is decreasing: absolute maximum f(1)=1f(1)=1, absolute minimum f(3)=13f(3)=\frac13.

(b) f(x)=1/xf(x)=1/x on (0,3](0,3]. The interval is not closed (it omits the left endpoint), so EVT does not apply. And indeed the maximum fails: as x→0+x\to 0^{+}, 1x→∞\dfrac1x\to\infty, so ff is unbounded above and has no absolute maximum. The minimum does still exist — ff is decreasing, so its smallest value is f(3)=13f(3)=\frac13. This shows EVT is sufficient, not necessary: losing a hypothesis loses the guarantee, not necessarily the conclusion.

(c) f(x)=x2f(x)=x^{2} on (−1,1)(-1,1). The interval is not closed, so EVT does not apply. There is no absolute maximum: f(x)<1f(x)<1 for every xx in (−1,1)(-1,1), but values get arbitrarily close to 11, so no single point achieves a largest value. The absolute minimum does exist: f(0)=0f(0)=0.

(d) The piecewise ff on [0,2][0,2]. The interval is closed and bounded, but ff is not continuous: lim⁡x→2−f(x)=4\lim_{x\to 2^{-}}f(x)=4 while f(2)=0f(2)=0. EVT does not apply. There is no absolute maximum: f(x)=x2<4f(x)=x^{2}<4 for 0≤x<20\le x<2 and f(2)=0f(2)=0, so values approach 44 without reaching it. The absolute minimum is 00, attained at both x=0x=0 and x=2x=2.

Summary: EVT applies only in (a). In (b), (c) and (d) the maximum genuinely fails to exist, while a minimum survives in all three by accident of the particular function.

State Fermat's Theorem with all of its hypotheses. Then give explicit counterexamples showing that each of the following statements is false:

(a) If f′(c)=0f'(c)=0 then ff has a local extremum at cc.

(b) If ff has a local extremum at cc then f′(c)=0f'(c)=0.

(c) If ff has an absolute extremum at cc on [a,b][a,b] then f′(c)=0f'(c)=0.

Show hint

Fermat needs three things: an extremum, an interior point, and an existing derivative. Each false statement below is obtained by deleting one of them (or by reversing the implication).

Show answer

Fermat's Theorem. If ff has a local maximum or a local minimum at cc, and cc is an interior point of the domain of ff, and f′(c)f'(c) exists, then f′(c)=0f'(c)=0.

(a) is the converse, and it is false. Take f(x)=x3f(x)=x^{3} and c=0c=0. Then f′(x)=3x2f'(x)=3x^{2}, so f′(0)=0f'(0)=0. But f′(x)=3x2>0f'(x)=3x^{2}>0 for every x≠0x\ne 0, so ff is strictly increasing on every interval containing 00: for any h>0h>0 we have f(−h)=−h3<0=f(0)<h3=f(h)f(-h)=-h^{3}<0=f(0)<h^{3}=f(h). So f(0)f(0) is neither ≥\ge nor ≤\le all nearby values, and 00 is not a local extremum. A stationary point need not be an extremum.

(b) is false because the derivative may not exist. Take f(x)=∣x∣f(x)=|x| and c=0c=0. Then f(0)=0≤∣x∣=f(x)f(0)=0\le|x|=f(x) for all xx, so ff has a local (in fact absolute) minimum at 00. But

lim⁡h→0+∣h∣−0h=1andlim⁡h→0−∣h∣−0h=−1,\lim_{h\to 0^{+}}\frac{|h|-0}{h}=1 \qquad\text{and}\qquad \lim_{h\to 0^{-}}\frac{|h|-0}{h}=-1,

so f′(0)f'(0) does not exist. Fermat's differentiability hypothesis is essential.

(c) is false because of endpoints. Take f(x)=xf(x)=x on [0,1][0,1]. Its absolute maximum on [0,1][0,1] is f(1)=1f(1)=1, at c=1c=1; but f′(1)=1≠0f'(1)=1\ne 0. The point c=1c=1 is not interior to [0,1][0,1], so Fermat says nothing. This is exactly why the Closed Interval Method tests the endpoints separately.

Correct use of Fermat (its contrapositive): if ff is continuous on [a,b][a,b], then every absolute extremum occurs either at a critical number in (a,b)(a,b) or at an endpoint. Fermat narrows the search; it never certifies that a candidate really is an extremum.

(a) Verify that f(x)=x3−3x2+2xf(x)=x^{3}-3x^{2}+2x satisfies the hypotheses of Rolle's Theorem on [0,2][0,2], and find all numbers cc the theorem provides.

(b) The function h(x)=1−x2/3h(x)=1-x^{2/3} satisfies h(−1)=h(1)h(-1)=h(1) and is continuous on [−1,1][-1,1], yet there is no cc in (−1,1)(-1,1) with h′(c)=0h'(c)=0. Explain why this does not contradict Rolle's Theorem.

Show hint

For (a), check the three hypotheses in order and then solve a quadratic — expect more than one answer. For (b), work out h′h' and look carefully at where it exists.

Show answer

(a) Hypotheses. ff is a polynomial, so it is continuous on [0,2][0,2] and differentiable on (0,2)(0,2). And

f(0)=0,f(2)=8−12+4=0,f(0)=0, \qquad f(2)=8-12+4=0,

so f(0)=f(2)f(0)=f(2). All three hypotheses of Rolle's Theorem hold.

Finding cc.

f′(x)=3x2−6x+2.f'(x)=3x^{2}-6x+2 .

Solve 3c2−6c+2=03c^{2}-6c+2=0 by the quadratic formula:

c=6±36−4(3)(2)2(3)=6±126=6±236=1±33.\begin{aligned} c &= \frac{6\pm\sqrt{36-4(3)(2)}}{2(3)} \\ &= \frac{6\pm\sqrt{12}}{6} \\ &= \frac{6\pm 2\sqrt3}{6} \\ &= 1\pm\frac{\sqrt3}{3} . \end{aligned}

Numerically 33≈0.5774\dfrac{\sqrt3}{3}\approx 0.5774, so c≈0.4226c\approx 0.4226 and c≈1.5774c\approx 1.5774. Both lie in (0,2)(0,2), so both are valid, illustrating that the cc in Rolle's Theorem need not be unique.

Check: f′(1)=3−6+2=−1<0f'(1)=3-6+2=-1<0 while f′(0)=2>0f'(0)=2>0 and f′(2)=12−12+2=2>0f'(2)=12-12+2=2>0, so f′f' does change sign twice on (0,2)(0,2) ✔.

(b) Differentiate:

h′(x)=−23x−1/3=−23x1/3.h'(x)=-\frac{2}{3}x^{-1/3}=-\frac{2}{3x^{1/3}} .

The numerator −23-\frac23 is never zero, so h′(x)≠0h'(x)\ne 0 for every xx where it is defined; and h′h' does not exist at x=0x=0, which lies in the open interval (−1,1)(-1,1).

So hh fails the hypothesis "differentiable on (−1,1)(-1,1)". Rolle's Theorem therefore makes no claim about hh, and no contradiction arises. Geometrically, hh has a sharp cusp peak at (0,1)(0,1): it does have a maximum there, but the graph turns without ever having a horizontal tangent.

Verify that f(x)=xx+2f(x)=\dfrac{x}{x+2} satisfies the hypotheses of the Mean Value Theorem on [1,4][1,4], and find the exact value of every cc that the theorem guarantees.

Show hint

Confirm that the point of discontinuity of this rational function lies outside [1,4][1,4], then set the derivative equal to the slope of the secant line and solve — watching which root actually lies in the open interval.

Show answer

Hypotheses. ff is a rational function; it is undefined only at x=−2x=-2, which is not in [1,4][1,4]. Hence ff is continuous on [1,4][1,4] and differentiable on (1,4)(1,4). MVT applies.

Secant slope.

f(1)=13,f(4)=46=23,f(1)=\frac{1}{3}, \qquad f(4)=\frac{4}{6}=\frac{2}{3},

f(4)−f(1)4−1=23−133=133=19.\frac{f(4)-f(1)}{4-1}=\frac{\frac23-\frac13}{3}=\frac{\frac13}{3}=\frac{1}{9} .

Derivative (quotient rule):

f′(x)=(1)(x+2)−(x)(1)(x+2)2=x+2−x(x+2)2=2(x+2)2.\begin{aligned} f'(x) &= \frac{(1)(x+2)-(x)(1)}{(x+2)^{2}} \\ &= \frac{x+2-x}{(x+2)^{2}} \\ &= \frac{2}{(x+2)^{2}} . \end{aligned}

Solve f′(c)=19f'(c)=\frac19.

2(c+2)2=19(c+2)2=18c+2=±32c=−2±32.\begin{aligned} \frac{2}{(c+2)^{2}} &= \frac{1}{9} \\ (c+2)^{2} &= 18 \\ c+2 &= \pm 3\sqrt2 \\ c &= -2\pm 3\sqrt2 . \end{aligned}

The root c=−2−32≈−6.24c=-2-3\sqrt2\approx -6.24 is not in (1,4)(1,4) and is rejected. The root

c=−2+32≈−2+4.2426=2.2426c=-2+3\sqrt2\approx -2+4.2426=2.2426

does lie in (1,4)(1,4).

Check: (c+2)2=(32)2=18(c+2)^{2}=(3\sqrt2)^{2}=18, so f′(c)=218=19f'(c)=\dfrac{2}{18}=\dfrac19 ✔, matching the secant slope.

Answer: c=32−2≈2.243c=3\sqrt2-2\approx 2.243, the only such value in (1,4)(1,4).

Suppose ff is continuous on [2,6][2,6] and differentiable on (2,6)(2,6), with f(2)=5f(2)=5 and 2≤f′(x)≤42\le f'(x)\le 4 for every xx in (2,6)(2,6). Determine the smallest and largest possible values of f(6)f(6), and show that each bound can actually be achieved.

Show hint

Apply the MVT to ff on the whole interval [2,6][2,6] and write the conclusion in the form f(6)=f(2)+f′(c)(6−2)f(6)=f(2)+f'(c)(6-2).

Show answer

Apply MVT on [2,6][2,6]. The hypotheses are given, so there is some cc with 2<c<62<c<6 and

f(6)−f(2)6−2=f′(c),i.e.f(6)−f(2)=4f′(c).\frac{f(6)-f(2)}{6-2}=f'(c), \qquad\text{i.e.}\qquad f(6)-f(2)=4f'(c).

Insert the bound on f′f'. Since cc lies in (2,6)(2,6), we know 2≤f′(c)≤42\le f'(c)\le 4. Multiplying the inequality by the positive number 44:

8≤4f′(c)≤16,so8≤f(6)−f(2)≤16.8\le 4f'(c)\le 16, \qquad\text{so}\qquad 8\le f(6)-f(2)\le 16 .

Use f(2)=5f(2)=5. Adding 55 throughout:

13≤f(6)≤21.13\le f(6)\le 21 .

Both bounds are attained.

  • f(x)=5+2(x−2)f(x)=5+2(x-2) is continuous and differentiable with f′(x)=2f'(x)=2 (inside the allowed range) and f(2)=5f(2)=5; it gives f(6)=5+2(4)=13f(6)=5+2(4)=13.
  • f(x)=5+4(x−2)f(x)=5+4(x-2) has f′(x)=4f'(x)=4 and f(2)=5f(2)=5; it gives f(6)=5+4(4)=21f(6)=5+4(4)=21.

Answer: the smallest possible value of f(6)f(6) is 1313 and the largest is 2121; that is, 13≤f(6)≤2113\le f(6)\le 21, and every value in that range occurs (for instance f(x)=5+m(x−2)f(x)=5+m(x-2) with 2≤m≤42\le m\le 4 produces f(6)=5+4mf(6)=5+4m, which sweeps out all of [13,21][13,21]).

Use the Mean Value Theorem to prove that 1+x<1+x2\sqrt{1+x}<1+\dfrac{x}{2} for every x>0x>0.

Show hint

Apply the MVT to f(t)=1+tf(t)=\sqrt{1+t} on the interval [0,x][0,x], then bound f′(c)f'(c) using the fact that cc is strictly positive.

Show answer

Fix x>0x>0 and let f(t)=1+t=(1+t)1/2f(t)=\sqrt{1+t}=(1+t)^{1/2}.

Hypotheses. On [0,x][0,x] we have 1+t≥1>01+t\ge 1>0, so ff is continuous on [0,x][0,x]; and

f′(t)=12(1+t)−1/2=121+tf'(t)=\frac{1}{2}(1+t)^{-1/2}=\frac{1}{2\sqrt{1+t}}

exists for every tt in (0,x)(0,x) (the denominator is never 00 there). MVT applies.

Apply MVT. There is a number cc with 0<c<x0<c<x such that

f(x)−f(0)=f′(c)(x−0),f(x)-f(0)=f'(c)(x-0),

that is,

1+x−1=x21+c.\sqrt{1+x}-1=\frac{x}{2\sqrt{1+c}} .

Bound the right-hand side. Because c>0c>0 we have 1+c>11+c>1, hence 1+c>1\sqrt{1+c}>1, hence 21+c>22\sqrt{1+c}>2. Dividing the positive number xx by a larger positive denominator makes the quotient smaller:

x21+c<x2.\frac{x}{2\sqrt{1+c}}<\frac{x}{2} .

Conclude.

1+x−1<x2  ⟹  1+x<1+x2for all x>0.\sqrt{1+x}-1<\frac{x}{2} \qquad\implies\qquad \sqrt{1+x}<1+\frac{x}{2} \quad\text{for all } x>0 .

Sanity checks. At x=3x=3: 4=2\sqrt{4}=2 versus 1+1.5=2.51+1.5=2.5 ✔. At x=0.21x=0.21: 1.21=1.1\sqrt{1.21}=1.1 versus 1+0.105=1.1051+0.105=1.105 ✔. Note the strictness genuinely requires x>0x>0; at x=0x=0 both sides equal 11, so the inequality becomes an equality there.

Let F(x)=arctan⁡x+arctan⁡1xF(x)=\arctan x+\arctan\dfrac{1}{x}, defined for all x≠0x\ne 0.

(a) Show that F′(x)=0F'(x)=0 for every x≠0x\ne 0.

(b) Explain why you may not conclude that FF is constant on its whole domain, and find the exact value of FF on each piece of the domain.

Show hint

Differentiate using ddxarctan⁡u=u′1+u2\dfrac{d}{dx}\arctan u=\dfrac{u'}{1+u^{2}} and simplify the compound fraction fully. Then re-read the hypothesis of the corollary "zero derivative implies constant" — what kind of set must the domain be?

Show answer

(a) Differentiate. With ddxarctan⁡u=u′1+u2\dfrac{d}{dx}\arctan u=\dfrac{u'}{1+u^{2}} and u=1xu=\dfrac1x, so u′=−1x2u'=-\dfrac{1}{x^{2}}:

F′(x)=11+x2+11+1x2⋅(−1x2)=11+x2+x2x2+1⋅(−1x2)=11+x2−1x2+1=0,\begin{aligned} F'(x) &= \frac{1}{1+x^{2}}+\frac{1}{1+\dfrac{1}{x^{2}}}\cdot\left(-\frac{1}{x^{2}}\right) \\ &= \frac{1}{1+x^{2}}+\frac{x^{2}}{x^{2}+1}\cdot\left(-\frac{1}{x^{2}}\right) \\ &= \frac{1}{1+x^{2}}-\frac{1}{x^{2}+1} \\ &= 0 , \end{aligned}

for every x≠0x\ne 0. (The middle step used 1+1x2=x2+1x21+\dfrac{1}{x^{2}}=\dfrac{x^{2}+1}{x^{2}}, whose reciprocal is x2x2+1\dfrac{x^{2}}{x^{2}+1}.)

(b) Why "constant" needs an interval. The corollary of the MVT states: if f′=0f'=0 on an interval II, then ff is constant on II. Its proof applies the MVT to [x1,x2][x_1,x_2], which requires the whole segment between x1x_1 and x2x_2 to lie in the domain. The domain of FF is the union of (−∞,0)(-\infty,0) and (0,∞)(0,\infty), which is not an interval: the segment from −1-1 to 11 leaves the domain at 00. So we may only conclude that FF is constant on each of the two intervals separately, possibly with different constants.

Evaluate one point in each piece.

For x>0x>0, take x=1x=1:

F(1)=arctan⁡1+arctan⁡1=π4+π4=π2.F(1)=\arctan 1+\arctan 1=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2} .

For x<0x<0, take x=−1x=-1:

F(−1)=arctan⁡(−1)+arctan⁡(−1)=−π4−π4=−π2.F(-1)=\arctan(-1)+\arctan(-1)=-\frac{\pi}{4}-\frac{\pi}{4}=-\frac{\pi}{2} .

Conclusion.

F(x)={π2,x>0−π2,x<0F(x)=\begin{cases}\dfrac{\pi}{2}, & x>0 \\ -\dfrac{\pi}{2}, & x<0\end{cases}

so FF takes two different values and is certainly not constant on its domain — no contradiction, because the domain is not an interval.

Numerical check: at x=2x=2, arctan⁡2≈1.10715\arctan 2\approx 1.10715 and arctan⁡0.5≈0.46365\arctan 0.5\approx 0.46365, summing to 1.57080≈π21.57080\approx\dfrac{\pi}{2} ✔; at x=−2x=-2 the sum is −1.57080-1.57080 ✔.

Let cc be any constant. Prove that the equation x3−15x+c=0x^{3}-15x+c=0 has at most one root in the interval [−2,2][-2,2].

Show hint

Argue by contradiction: assume there are two roots, and ask what Rolle's Theorem would then force to exist. Then check whether such a point can live inside (−2,2)(-2,2).

Show answer

Let f(x)=x3−15x+cf(x)=x^{3}-15x+c, a polynomial, so ff is continuous and differentiable on every real number — in particular on any closed subinterval of [−2,2][-2,2].

Suppose, for contradiction, that ff has two distinct roots in [−2,2][-2,2], say a<ba<b with f(a)=f(b)=0f(a)=f(b)=0.

Apply Rolle's Theorem on [a,b][a,b]. The hypotheses hold: ff is continuous on [a,b][a,b], differentiable on (a,b)(a,b), and f(a)=f(b)=0f(a)=f(b)=0. Therefore there exists rr with a<r<ba<r<b such that

f′(r)=0.f'(r)=0 .

Since −2≤a<r<b≤2-2\le a<r<b\le 2, this rr lies in the open interval (−2,2)(-2,2).

But f′f' has no zero there.

f′(x)=3x2−15=0  ⟹  x2=5  ⟹  x=±5.f'(x)=3x^{2}-15=0 \quad\implies\quad x^{2}=5 \quad\implies\quad x=\pm\sqrt5 .

Since 5≈2.236>2\sqrt5\approx 2.236>2, neither 5\sqrt5 nor −5-\sqrt5 lies in (−2,2)(-2,2). So no such rr exists — a contradiction.

Therefore ff cannot have two distinct roots in [−2,2][-2,2]: it has at most one.

Independent confirmation. For −2<x<2-2<x<2 we have x2<4<5x^{2}<4<5, so f′(x)=3x2−15<3(5)−15=0f'(x)=3x^{2}-15<3(5)-15=0. A function with a strictly negative derivative throughout an interval is strictly decreasing there, and a strictly decreasing function takes each value at most once — the same conclusion, reached without contradiction.

Note the wording "at most one". The statement does not promise a root exists. For c=100c=100, ff is decreasing on [−2,2][-2,2] with minimum value f(2)=8−30+100=78>0f(2)=8-30+100=78>0, so there is no root at all. For c=0c=0, however, f(x)=x(x2−15)f(x)=x\left(x^{2}-15\right) has the single root x=0x=0 in [−2,2][-2,2] (the others, ±15≈±3.873\pm\sqrt{15}\approx\pm 3.873, fall outside).