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Applications of the Derivative · Topic 16 of 23

Monotonicity, Concavity and the Derivative Tests

The sign of f′f' says where the graph rises or falls; the sign of f′′f'' says which way it bends. Build one sign chart for each: a sign change of f′f' at a critical number is a local maximum or minimum, and a sign change of f′′f'' at a point of the graph is an inflection point.

5 min readCore — on every final7 formulas6 quick checks
01

Key ideas

5 things to remember
  1. 1

    The sign of f′f' gives direction

    f′>0f'>0 on an interval means ff is increasing there; f′<0f'<0 means decreasing. Monotonicity is a statement about an interval, so report the intervals one at a time — never joined with ∪\cup.

  2. 2

    Critical numbers are only candidates

    cc is critical when cc is in the domain of ff and f′(c)=0f'(c)=0 or f′(c)f'(c) does not exist. Every interior extremum sits at a critical number, but x3x^3 at 00 shows the converse fails.

  3. 3

    First Derivative Test: the sign must change

    At a critical number cc where ff is continuous: f′f' going ++ to −- is a local maximum, −- to ++ a local minimum, and no change means no extremum. It still works at corners and cusps.

  4. 4

    f′′f'' bends; inflection is a sign change

    f′′>0f''>0 is concave up, f′′<0f''<0 concave down; an inflection point is where the concavity actually changes at a point of the graph. Bending says nothing about direction: 1/x1/x falls while concave up on (0,∞)(0,\infty).

  5. 5

    Second Derivative Test can say nothing

    With f′(c)=0f'(c)=0: f′′(c)>0f''(c)>0 gives a local minimum, f′′(c)<0f''(c)<0 a local maximum, and f′′(c)=0f''(c)=0 gives no information at all — x4x^4, −x4-x^4 and x3x^3 all land there. Fall back on the First Derivative Test.

-2-112-4-224xylocal max (−1, 2)local min (1, −2)inflection (0, 0)f(x) = x³ − 3xf′ > 0f′ < 0f′ > 0f″ < 0f″ > 0
f′(x)=3x2−3f'(x)=3x^2-3 is positive on the shaded bands, so ff increases on (−∞,−1](-\infty,-1] and [1,∞)[1,\infty) and decreases on [−1,1][-1,1]; the sign flips at each extremum. f′′(x)=6xf''(x)=6x flips sign at x=0x=0, so (0,0)(0,0) is the inflection point — concave down to its left, up to its right.
rising, steepeningrising, levelling offfalling, levelling offfalling, steepeningf′ > 0, f″ > 0f′ > 0, f″ < 0f′ < 0, f″ > 0f′ < 0, f″ < 0
All four combinations occur, so the two signs must be read separately: f′f' decides up or down, f′′f'' decides the bend. Models on (0,∞)(0,\infty): x2x^2, x\sqrt{x}, 1/x1/x, −x2-x^2. A concave-up graph can perfectly well be falling.
02

Formulas

What to have memorised

  • Increasing / decreasing test

    f′>0 on I⇒f increasing on I;f′<0⇒decreasingf'>0 \text{ on } I \Rightarrow f \text{ increasing on } I;\quad f'<0 \Rightarrow \text{decreasing}

    II must be an interval — a union across a gap in the domain proves nothing.

  • Critical number

    f′(c)=0  or  f′(c) undefined,c∈domain(f)f'(c)=0 \ \text{ or }\ f'(c)\ \text{undefined},\qquad c\in\text{domain}(f)

    Candidates only: every interior extremum is critical, not every critical number is an extremum.

  • First Derivative Test

    f′: +→− at c ⇒ local max;−→+ ⇒ local minf':\ +\to-\ \text{at } c\ \Rightarrow\ \text{local max};\qquad -\to+\ \Rightarrow\ \text{local min}

    No sign change means no extremum; ff must be continuous at cc.

  • Concavity test

    f′′>0 on I⇒f′ increasing, f concave up on If''>0 \text{ on } I \Rightarrow f' \text{ increasing},\ f \text{ concave up on } I

    Concave up is how the graph bends, not whether it rises.

  • Second Derivative Test

    f′(c)=0, f′′(c)>0⇒local min;f′′(c)<0⇒local maxf'(c)=0,\ f''(c)>0 \Rightarrow \text{local min};\qquad f''(c)<0 \Rightarrow \text{local max}

    f′′(c)=0f''(c)=0 gives no information — go back to the First Derivative Test.

  • Inflection point

    f′′ changes sign at c,c∈domain(f)f'' \text{ changes sign at } c,\qquad c\in\text{domain}(f)

    Candidates: f′′(c)=0f''(c)=0 or f′′(c)f''(c) undefined. A zero alone proves nothing.

  • Mean Value Theorem (the engine)

    f(b)−f(a)=f′(c) (b−a),c∈(a,b)f(b)-f(a)=f'(c)\,(b-a),\qquad c\in(a,b)

    Both sign tests are corollaries: the sign of f′(c)f'(c) fixes the sign of f(b)−f(a)f(b)-f(a).

03

Run the two sign charts

The steps, in order
  1. 1

    State the domain first and mark every excluded point on the number line — it splits the chart but is never a critical number.

  2. 2

    Differentiate, put f′f' over one denominator and factor completely. Critical numbers: f′(c)=0f'(c)=0 or f′(c)f'(c) undefined, with cc in the domain.

  3. 3

    Read the sign of f′f' on each open interval by multiplying factor signs. A factor to an even power never flips the sign.

  4. 4

    Classify each critical number by the change: ++ to −- local maximum, −- to ++ local minimum, no change no extremum.

  5. 5

    Repeat with f′′f'': candidates are where f′′=0f''=0 or f′′f'' is undefined; a genuine change of sign at a point of the graph is an inflection point.

  6. 6

    Report location and value separately: local maximum value f(c)f(c) at x=cx=c; inflection point (c,f(c))(c,f(c)).

04

Watch out

The mistakes that cost marks

  • ✗ Wrong

    f′(x)=(x−1)2(x−4)f'(x)=(x-1)^2(x-4) is zero at x=1x=1, so ff has an extremum there.

    ✓ Right

    (x−1)2≥0(x-1)^2\ge0 never flips the sign, so f′<0f'<0 on both sides of 11: no extremum.

    Why: f′(c)=0f'(c)=0 only makes cc a candidate; the sign has to change.

  • ✗ Wrong

    f′′>0f''>0, so ff is increasing.

    ✓ Right

    f′′>0f''>0 says f′f' is increasing — the graph is concave up, rising or not.

    Why: x2x^2 on (−∞,0)(-\infty,0) is concave up and decreasing.

  • ✗ Wrong

    f′′(0)=0f''(0)=0, so (0,f(0))(0,f(0)) is an inflection point.

    ✓ Right

    Check that f′′f'' changes sign at 00 before calling it one.

    Why: f′′(x)=12x2≥0f''(x)=12x^2\ge0 for x4x^4: concave up on both sides, no inflection.

  • ✗ Wrong

    f′′(c)=0f''(c)=0, so the Second Derivative Test says cc is not an extremum.

    ✓ Right

    The test says nothing at all; classify cc with the First Derivative Test.

    Why: x4x^4, −x4-x^4 and x3x^3 all have f′(0)=f′′(0)=0f'(0)=f''(0)=0 with three different outcomes.

  • ✗ Wrong

    x=2x=2 is a critical number of f(x)=1x−2f(x)=\dfrac{1}{x-2}.

    ✓ Right

    22 is not in the domain: it splits the sign chart but is never critical, and never an inflection point.

  • ✗ Wrong

    ff is increasing on (−∞,0)∪(0,∞)(-\infty,0)\cup(0,\infty).

    ✓ Right

    ff is increasing on (−∞,0)(-\infty,0) and increasing on (0,∞)(0,\infty) — two statements.

    Why: For f(x)=−1/xf(x)=-1/x the union claim is false: f(−1)=1>−1=f(1)f(-1)=1>-1=f(1).

05

Quick check

Commit to an answer before you reveal one
  1. Q1easy

    Let f(x)=2x3−3x2−12x+4f(x)=2x^3-3x^2-12x+4. Find the intervals of increase and decrease, and use the First Derivative Test to locate and classify every local extremum.

    Hint

    Factor the derivative completely before you read any signs.

    Show answer

    Answer

    Increasing on (−∞,−1](-\infty,-1] and [2,∞)[2,\infty), decreasing on [−1,2][-1,2]; local maximum value 1111 at x=−1x=-1, local minimum value −16-16 at x=2x=2.

    Steps

    f′(x)=6x2−6x−12=6(x−2)(x+1)f'(x)=6x^2-6x-12=6(x-2)(x+1), so the critical numbers are x=−1x=-1 and x=2x=2.

    Signs: f′(−2)=24>0f'(-2)=24>0, f′(0)=−12<0f'(0)=-12<0, f′(3)=24>0f'(3)=24>0 — so f′f' is +,−,++,-,+ across (−∞,−1)(-\infty,-1), (−1,2)(-1,2), (2,∞)(2,\infty).

    ++ to −- at x=−1x=-1: local maximum, f(−1)=−2−3+12+4=11f(-1)=-2-3+12+4=11. −- to ++ at x=2x=2: local minimum, f(2)=16−12−24+4=−16f(2)=16-12-24+4=-16.

    Check with f′′(x)=12x−6f''(x)=12x-6: f′′(−1)=−18<0f''(-1)=-18<0 and f′′(2)=18>0f''(2)=18>0.

  2. Q2easy

    Determine the intervals of concavity of f(x)=x3−6x2+9x+1f(x)=x^3-6x^2+9x+1 and give the coordinates of every inflection point.

    Hint

    Concavity is controlled by f′′f'', not f′f' — and the sign must actually change.

    Show answer

    Answer

    Concave down on (−∞,2)(-\infty,2), concave up on (2,∞)(2,\infty); the only inflection point is (2,3)(2,3).

    Steps

    f′(x)=3x2−12x+9f'(x)=3x^2-12x+9 and f′′(x)=6x−12=6(x−2)f''(x)=6x-12=6(x-2), so x=2x=2 is the only candidate.

    f′′(0)=−12<0f''(0)=-12<0 and f′′(3)=6>0f''(3)=6>0: concave down on (−∞,2)(-\infty,2), concave up on (2,∞)(2,\infty). The sign genuinely changes and ff is continuous, so x=2x=2 gives an inflection point, with f(2)=8−24+18+1=3f(2)=8-24+18+1=3.

    Note f′(2)=−3≠0f'(2)=-3\neq0: an inflection point needs no horizontal tangent, and x=2x=2 is not a critical number.

  3. Q3medium

    Let f(x)=3x5−5x3f(x)=3x^5-5x^3. Find every critical number and classify it, using the Second Derivative Test where it works and the First Derivative Test where it does not. Then find all inflection points.

    Hint

    One critical number makes f′′f'' vanish — that is a signal to switch tests, not an answer.

    Show answer

    Answer

    Local maximum 22 at x=−1x=-1; local minimum −2-2 at x=1x=1; no extremum at x=0x=0. Inflection points (−22,728)\left(-\frac{\sqrt2}{2},\frac{7\sqrt2}{8}\right), (0,0)(0,0) and (22,−728)\left(\frac{\sqrt2}{2},-\frac{7\sqrt2}{8}\right).

    Steps

    f′(x)=15x4−15x2=15x2(x−1)(x+1)f'(x)=15x^4-15x^2=15x^2(x-1)(x+1) and f′′(x)=30x(2x2−1)f''(x)=30x(2x^2-1). Critical numbers: −1,0,1-1,0,1.

    f′′(−1)=−30<0f''(-1)=-30<0: local maximum f(−1)=2f(-1)=2. f′′(1)=30>0f''(1)=30>0: local minimum f(1)=−2f(1)=-2. f′′(0)=0f''(0)=0 is inconclusive, so use f′f': 15x2(x2−1)<015x^2(x^2-1)<0 on (−1,0)(-1,0) and on (0,1)(0,1) — no sign change, no extremum at 00.

    f′′=0f''=0 at x=0,±22x=0,\pm\frac{\sqrt2}{2} and changes sign at all three, so all three are inflection points: f(22)=328−1028=−728f\left(\frac{\sqrt2}{2}\right)=\frac{3\sqrt2}{8}-\frac{10\sqrt2}{8}=-\frac{7\sqrt2}{8}, and ff is odd.

  4. Q4medium

    Let f(x)=x−1(x+2)2f(x)=\dfrac{x-1}{(x+2)^2}. State the domain, then find the intervals of increase and decrease, every local extremum, the intervals of concavity and every inflection point.

    Hint

    Mark the excluded point first: (x+2)3(x+2)^3 changes sign there, (x+2)4(x+2)^4 does not.

    Show answer

    Answer

    Domain x≠−2x\neq-2; decreasing on (−∞,−2)(-\infty,-2) and [4,∞)[4,\infty), increasing on (−2,4](-2,4]; local maximum value 112\frac{1}{12} at x=4x=4 and no local minimum; concave down on (−∞,−2)(-\infty,-2) and (−2,7)(-2,7), concave up on (7,∞)(7,\infty); inflection point (7,227)\left(7,\frac{2}{27}\right).

    Steps

    Domain: x≠−2x\neq-2. That point splits every chart but is not a critical number.

    f′(x)=(x+2)−2(x−1)(x+2)3=4−x(x+2)3f'(x)=\frac{(x+2)-2(x-1)}{(x+2)^3}=\frac{4-x}{(x+2)^3}

    The only critical number is x=4x=4. Signs: f′(−3)=−7<0f'(-3)=-7<0, f′(0)=12>0f'(0)=\frac12>0, f′(5)<0f'(5)<0. So ++ to −- at 44: local maximum f(4)=336=112f(4)=\frac{3}{36}=\frac{1}{12}.

    f′′(x)=−(x+2)−3(4−x)(x+2)4=2(x−7)(x+2)4f''(x)=\frac{-(x+2)-3(4-x)}{(x+2)^4}=\frac{2(x-7)}{(x+2)^4}

    (x+2)4>0(x+2)^4>0, so f′′f'' has the sign of x−7x-7, and f(7)=681=227f(7)=\frac{6}{81}=\frac{2}{27}.

  5. Q5medium

    Find constants aa and bb so that f(x)=x3+ax2+bx+1f(x)=x^3+ax^2+bx+1 has a local maximum at x=−1x=-1 and an inflection point at x=1x=1, then verify that your ff really has both features.

    Hint

    Each requirement becomes an equation; both are only necessary conditions, so the check is compulsory.

    Show answer

    Answer

    a=−3a=-3, b=−9b=-9, giving f(x)=x3−3x2−9x+1f(x)=x^3-3x^2-9x+1: local maximum value 66 at x=−1x=-1, inflection point (1,−10)(1,-10).

    Steps

    f′(x)=3x2+2ax+bf'(x)=3x^2+2ax+b and f′′(x)=6x+2af''(x)=6x+2a.

    Inflection at x=1x=1 forces f′′(1)=6+2a=0f''(1)=6+2a=0, so a=−3a=-3. A local maximum at the interior point x=−1x=-1 forces f′(−1)=3−2a+b=0f'(-1)=3-2a+b=0, so 9+b=09+b=0 and b=−9b=-9.

    Verify: f′(x)=3(x−3)(x+1)f'(x)=3(x-3)(x+1) and f′′(x)=6(x−1)f''(x)=6(x-1). Then f′(−1)=0f'(-1)=0 with f′′(−1)=−12<0f''(-1)=-12<0, a local maximum of value f(−1)=6f(-1)=6; and f′′f'' changes sign at 11, so (1,−10)(1,-10) is an inflection point.

  6. Q6hard

    ff is twice differentiable on R\mathbb{R} with f′(x)=(x−1)2[(x−1)2−9]f'(x)=(x-1)^2\left[(x-1)^2-9\right].

    (a) Find the critical numbers and the intervals of increase and decrease, and classify each critical number. (b) Find every xx at which ff has an inflection point. (c) Why can you not give the yy-coordinates?

    Hint

    Factor the bracket as a difference of squares; for (b) put u=x−1u=x-1.

    Show answer

    Answer

    (a) Critical numbers −2,1,4-2,1,4; increasing on (−∞,−2](-\infty,-2] and [4,∞)[4,\infty), decreasing on [−2,4][-2,4]; local maximum at x=−2x=-2, local minimum at x=4x=4, neither at x=1x=1. (b) x=1x=1 and x=1±322x=1\pm\frac{3\sqrt2}{2}. (c) ff is determined only up to an additive constant.

    Steps

    (a) (x−1)2−9=(x−4)(x+2)(x-1)^2-9=(x-4)(x+2), so f′(x)=(x−1)2(x−4)(x+2)f'(x)=(x-1)^2(x-4)(x+2) and the critical numbers are −2,1,4-2,1,4. The even factor never flips the sign, so f′f' has the sign of (x−4)(x+2)(x-4)(x+2): +,−,++,-,+. Hence a local maximum at −2-2, a local minimum at 44, and only a horizontal tangent at 11.

    (b) With u=x−1u=x-1, f′=u4−9u2f'=u^4-9u^2, so f′′=4u3−18u=2u(2u2−9)f''=4u^3-18u=2u(2u^2-9), zero at u=0,±322u=0,\pm\frac{3\sqrt2}{2} — that is x=1x=1 and x=1±322x=1\pm\frac{3\sqrt2}{2} — and each is a genuine sign change.

    (c) Every f+Cf+C has the same f′f'.

06

On the exam

How this topic is marked

  • Marks are for the chart, not the verdict: show the factored f′f', the split points including any excluded from the domain, the sign on each interval, then the conclusion in words.

  • Keep location and value apart — write "local maximum value 1111 at x=−1x=-1" and "inflection point (2,3)(2,3)", never "maximum at y=11y=11".

  • If f′′(c)=0f''(c)=0 at a critical number, do not write "no extremum": say the Second Derivative Test is inconclusive and finish with the First Derivative Test.

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