Monotonicity, Concavity and the Derivative Tests

The derivative tells you the slope of a graph at a point; this topic converts sign information about $f'$ and $f''$ into shape information about the graph of $f$ — where it rises, where it falls, which way it bends, and which critical points are peaks and which are valleys. Every theorem below is a corollary of the Mean Value Theorem, and everything that follows in the course (curve sketching, optimisation) is an application of these tests. The recurring discipline is the sign chart: factor a derivative, mark the points where it can change sign, and read off the geometry.

Vocabulary and standing definitions

Increasing / decreasing. $f$ is increasing on an interval $I$ if $x_1<x_2$ in $I$ implies $f(x_1)<f(x_2)$; decreasing if $x_1<x_2$ implies $f(x_1)>f(x_2)$. (The non-strict versions weaken the conclusion to $f(x_1)\le f(x_2)$ and $f(x_1)\ge f(x_2)$ respectively.) Monotonicity is a statement about an interval, never about a single point.

Critical number. $c$ is a critical number of $f$ if $c$ is in the domain of $f$ and either $f'(c)=0$ or $f'(c)$ does not exist. A point outside the domain — a vertical asymptote of a rational function, say — is never a critical number, even though it does split the number line.

Local extremum. $f$ has a local maximum at $c$ if $f(x)\le f(c)$ for all $x$ in some open interval containing $c$ (local minimum: $f(x)\ge f(c)$).

Fermat's theorem. If $c$ is an interior point of the domain of $f$, if $f$ has a local extremum at $c$, and if $f'(c)$ exists, then $f'(c)=0$. The interior-point hypothesis is not decoration: $f(x)=x$ on $[0,1]$ has an absolute minimum at the endpoint $c=0$, yet the derivative there (one-sided, the only kind available) equals $1$. Consequently every interior local extremum occurs at a critical number — but the converse is false, which is exactly why we need the tests below. On a closed-interval domain the two endpoints are never critical numbers and must be examined separately.

Monotonicity from the sign of $f'$

Mean Value Theorem (the engine behind everything below). If $f$ is continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$, then there is at least one $c\in(a,b)$ with $f'(c)=\frac{f(b)-f(a)}{b-a},\qquad\text{equivalently}\qquad f(b)-f(a)=f'(c)\,(b-a).$ Both hypotheses are required, and nothing at all is assumed about $f'$ at the endpoints $a$ and $b$.

Increasing / Decreasing Test. Let $f$ be continuous on an interval $I$ and differentiable at every interior point of $I$.

• If $f'(x)>0$ at every interior point, $f$ is increasing on $I$.
• If $f'(x)<0$ at every interior point, $f$ is decreasing on $I$.
• If $f'(x)=0$ at every interior point, $f$ is constant on $I$.

Proof from the MVT. Let $x_1<x_2$ lie in $I$. Then $f$ is continuous on $[x_1,x_2]$ and differentiable on $(x_1,x_2)$, so the Mean Value Theorem supplies a $c\in(x_1,x_2)$ with $f(x_2)-f(x_1)=f'(c)\,(x_2-x_1).$ Since $x_2-x_1>0$, the sign of $f(x_2)-f(x_1)$ is exactly the sign of $f'(c)$. If $f'>0$ throughout, then $f(x_2)>f(x_1)$ for every such pair, which is the definition of increasing.

Two things the proof makes visible:

1. $I$ must be an interval. For $f(x)=-\dfrac1x$ we have $f'(x)=\dfrac{1}{x^2}>0$ on the whole domain, yet $f(-1)=1>-1=f(1)$. The domain $(-\infty,0)\cup(0,\infty)$ is not an interval, so no MVT is available across the gap.
2. The converse fails. $f(x)=x^3$ is increasing on $\mathbb{R}$ but $f'(0)=0$. The sharp statement is: for $f$ differentiable on $I$, $f$ is increasing on $I$ exactly when $f'\ge0$ on $I$ and $f'$ vanishes identically on no subinterval of positive length. Isolated zeros of $f'$ never spoil monotonicity, so you may always close up the intervals: if $f'<0$ on $(a,c)$ and on $(c,b)$ and $f$ is continuous at $c$, then $f$ is decreasing on the whole of $(a,b)$.

Sign charts, done correctly

1. Domain first. Write down the domain of $f$ and mark every excluded point.
2. Differentiate, then put $f'$ over a single denominator and factor it completely. You cannot read the sign of $4x^3-12x^2$; you can read the sign of $4x^2(x-3)$.
3. Mark the split points: zeros of the numerator, zeros of the denominator, and the excluded points from step 1.
4. Determine the sign on each open interval by multiplying the signs of the factors, or by testing one convenient point.
5. Report interval by interval. A factor raised to an even power is $\ge0$ and therefore never flips the sign.

The First Derivative Test

Theorem. Let $c$ be a critical number of $f$, and suppose $f$ is continuous at $c$ and differentiable on a punctured interval $(c-\delta,c)\cup(c,c+\delta)$.

• $f'>0$ on $(c-\delta,c)$ and $f'<0$ on $(c,c+\delta)$ $\Rightarrow$ local maximum at $c$.
• $f'<0$ then $f'>0$ $\Rightarrow$ local minimum at $c$.
• $f'$ has the same sign on both sides $\Rightarrow$ no local extremum at $c$.

Continuity at $c$ is indispensable: a function with a jump at $c$ can have $f'$ change from $+$ to $-$ there and still have no local maximum. The test also requires $f'$ to keep a constant sign on each side; for $f(x)=x^2\big(2+\sin(1/x)\big)$ with $f(0)=0$ the derivative changes sign infinitely often near $0$, so the test does not apply even though $f$ clearly has a minimum at $0$ (because $f(x)\ge x^2>0$ for $x\ne0$).

Concavity

Definition. $f$ is concave up on an interval $I$ if, for every $a\in I$ at which $f$ is differentiable, the part of the graph lying over $I$ stays above the tangent line at $a$ and touches it only at $x=a$; equivalently — when $f$ is differentiable on $I$ — if $f'$ is increasing on $I$; equivalently (no differentiability required anywhere), if for every $a<b$ in $I$ the chord from $(a,f(a))$ to $(b,f(b))$ lies strictly above the graph on $(a,b)$. Concave down reverses all three. The words "over $I$" are load-bearing: a graph concave up on $I$ may perfectly well drop below one of those tangent lines at points outside $I$.

Concavity Test. If $f''(x)>0$ for all $x$ in an interval $I$, then $f$ is concave up on $I$; if $f''(x)<0$ on $I$, then $f$ is concave down on $I$.

Why: $f''>0$ says $f'$ is increasing (apply the Increasing/Decreasing Test to $f'$), and that is the definition. The converse fails: $f(x)=x^4$ is concave up on all of $\mathbb{R}$ although $f''(0)=0$.

Concavity is independent of monotonicity — all four combinations occur:

Inflection points

Definition. $(c,f(c))$ is an inflection point of $f$ if $f$ is continuous at $c$ and the concavity changes at $c$ — up on one side, down on the other.

Necessary condition. If $(c,f(c))$ is an inflection point and $f''(c)$ exists, then $f''(c)=0$.

So the candidates are the points $c$ in the domain of $f$ where $f''(c)=0$ or $f''(c)$ fails to exist; each candidate must then be tested for an actual change of sign of $f''$.

Why $f''(c)=0$ alone is not enough, in three examples:

• $f(x)=x^4$: $f''(0)=0$, but $f''(x)=12x^2\ge0$ everywhere, so the concavity never changes. $(0,0)$ is a minimum, not an inflection point.
• $f(x)=x^{1/3}$: $f'(x)=\tfrac13x^{-2/3}$ and $f''(x)=-\tfrac29x^{-5/3}$, which is never zero and does not exist at $0$. Yet $f''>0$ for $x<0$, $f''<0$ for $x>0$, and $f$ is continuous at $0$, so $(0,0)$ is an inflection point (with a vertical tangent). Candidates where $f''$ is undefined must not be skipped.
• $f(x)=1/x$: $f''(x)=2/x^3$ is negative for $x<0$ and positive for $x>0$, so the concavity does "change" across $0$ — but $0$ is not in the domain, there is no point $(0,f(0))$ on the graph, and there is no inflection point.

The Second Derivative Test

Theorem. Suppose $f'(c)=0$ and $f''(c)$ exists.

• $f''(c)>0$ $\Rightarrow$ $f$ has a local minimum at $c$.
• $f''(c)<0$ $\Rightarrow$ $f$ has a local maximum at $c$.
• $f''(c)=0$ $\Rightarrow$ the test gives no information whatsoever.

(The hypothesis that $f''(c)$ exists already forces $f'$ to exist on an open interval around $c$. Many textbooks additionally assume $f''$ is continuous near $c$; that is more than is needed.)

Why: since $f'(c)=0$, $f''(c)=\lim_{x\to c}\frac{f'(x)-f'(c)}{x-c}=\lim_{x\to c}\frac{f'(x)}{x-c}.$ If this limit is positive, then $\dfrac{f'(x)}{x-c}>0$ for all $x$ near $c$ (with $x\ne c$), which forces $f'(x)<0$ just to the left of $c$ and $f'(x)>0$ just to the right. The First Derivative Test then gives a local minimum.

The inconclusive case. At $c=0$ each of the following has $f'(0)=f''(0)=0$:

So $f''(c)=0$ means "try something else", never "no extremum". Fall back on the First Derivative Test. The Second Derivative Test is also silent at a critical number where $f'(c)$ does not exist (a corner or a cusp) — the theorem cannot even be stated there.

Worked example 1 — a full analysis of $f(x)=x^4-4x^3$

Step 1 — domain. $f$ is a polynomial: defined and twice differentiable on all of $\mathbb{R}$.

Step 2 — critical numbers. $f'(x)=4x^3-12x^2=4x^2(x-3).$ $f'$ exists everywhere, and $f'(x)=0$ exactly at $x=0$ and $x=3$. Those are the only critical numbers.

Step 3 — sign chart for $f'$. The factor $4x^2$ is $\ge0$ and vanishes only at $0$, so it never changes the sign; the sign of $f'$ is the sign of $x-3$.

Because $f$ is continuous at $0$ and $f'<0$ on both sides of it, $f$ is decreasing on the whole interval $(-\infty,3]$ and increasing on $[3,\infty)$.

Step 4 — First Derivative Test. At $x=0$ there is no sign change, so there is no extremum — only a horizontal tangent. At $x=3$ the sign changes from $-$ to $+$: a local minimum with value $f(3)=81-108=-27$.

Step 5 — concavity. $f''(x)=12x^2-24x=12x(x-2),$ so the candidates are $x=0$ and $x=2$.

Both candidates show a genuine change, so there are two inflection points: $(0,0)$ and $(2,-16)$.

Step 6 — cross-check with the Second Derivative Test. $f''(3)=12(9)-24(3)=108-72=36>0$, confirming the local minimum at $x=3$. At $x=0$, $f''(0)=0$, so the test is inconclusive — and indeed $x=0$ is not an extremum. That is the inconclusive case in the wild.

Worked example 2 — a critical number where $f'$ does not exist

Analyse $f(x)=x^{2/3}(x-5)=x^{5/3}-5x^{2/3}$, defined for all real $x$ (the exponents have odd denominator $3$, so $x^{2/3}=\left(x^{1/3}\right)^2$ makes sense for $x<0$).

Step 1 — first derivative over one denominator. $f'(x)=\frac53x^{2/3}-\frac{10}{3}x^{-1/3}=\frac{5x}{3x^{1/3}}-\frac{10}{3x^{1/3}}=\frac{5(x-2)}{3x^{1/3}}.$

Step 2 — critical numbers. $f'(x)=0$ at $x=2$. Also $f'(0)$ does not exist while $0$ is in the domain of $f$, so $x=0$ is a critical number too.

Step 3 — sign chart. $3x^{1/3}<0$ for $x<0$ and $>0$ for $x>0$.

First Derivative Test: $+$ to $-$ at $x=0$ gives a local maximum $f(0)=0$; $-$ to $+$ at $x=2$ gives a local minimum $f(2)=2^{2/3}(2-5)=-3\cdot2^{2/3}\approx-4.76$. The Second Derivative Test could not have decided $x=0$: $f'(0)$ does not exist. Since $f'(x)\to+\infty$ as $x\to0^-$ and $f'(x)\to-\infty$ as $x\to0^+$, the graph has a cusp at the origin.

Step 4 — concavity. $f''(x)=\frac{10}{9}x^{-1/3}+\frac{10}{9}x^{-4/3}=\frac{10}{9}\cdot\frac{x+1}{x^{4/3}}.$ For every $x\ne0$, $x^{4/3}=\left(x^{1/3}\right)^4>0$, so the sign of $f''$ is the sign of $x+1$: concave down on $(-\infty,-1)$, concave up on $(-1,0)$ and on $(0,\infty)$.

Step 5 — inflection points. The candidates were $x=-1$ (where $f''=0$) and $x=0$ (where $f''$ is undefined). Only $x=-1$ produces a change of concavity, and $f(-1)=(-1)^{2/3}(-1-5)=1\cdot(-6)=-6$, so the unique inflection point is $(-1,-6)$. At the cusp $x=0$ the graph is concave up on both sides, so it is not an inflection point.

Common mistakes

Quick reference