Grade 12

Applications of the Derivative

How the sign and zeros of the derivative reveal where a function increases, decreases, and reaches its extrema, culminating in an algorithm for global maxima/minima on an interval and real optimization problems.

IntuitionReading the shape of a curve from its slope

Picture the elevation profile of a hiking trail plotted as $f(x)$ , where $x$ is the distance walked. When you are climbing, the trail's slope $f'(x)$ is positive; when descending, $f'(x)$ is negative; and right at a summit or a valley bottom, the ground is momentarily flat, so $f'(x)=0$ . The whole theory of "monotonicity and extrema" is exactly this everyday observation turned into a precise, provable statement: the sign of the derivative tells you where a function rises and falls, and the zeros of the derivative mark the candidate turning points.

The cubic

f(x)=x^3-3x+2

. Drag the sliders

a,b,c,d

f(x)=ax^3+bx^2+cx+d

and watch how the intervals where the curve rises or falls, and the location of its "hilltop" and "valley", change together with the shape of the graph.

SchoolMonotonicity and extrema on an interval

Definition: Increasing, decreasing, and local extrema

A function $f$ is increasing on an interval $K$ if for all $x_1,x_2\in K$ with $x_1<x_2$ we have $f(x_1)<f(x_2)$ , and decreasing if instead $f(x_1)>f(x_2)$ . A point $x_0$ interior to the domain is a local maximum if there is a $\delta>0$ with $f(x)\le f(x_0)$ for every $x$ within $\delta$ of $x_0$ ; a local minimum is defined the same way with the inequality reversed. Points where $f'(x_0)=0$ or $f'(x_0)$ fails to exist are called critical points — the only candidates for local extrema.

f'(x) > 0 \;\forall\, x \in (a,b) \implies \forall\, x_1 < x_2 \in (a,b),\ f(x_1) < f(x_2)

Checking that inequality directly from the definition is awkward for a complicated $f$ . The next theorem replaces it with a test you can actually compute: look at the sign of $f'(x)$ .

f'(x_0)=0,\ f''(x_0)>0 \implies f(x)\ge f(x_0); \qquad f'(x_0)=0,\ f''(x_0)<0 \implies f(x)\le f(x_0)

Reading monotonicity, extrema, and concavity from the derivatives
Sign condition	Conclusion
$f'(x)>0$ on $(a,b)$	$f$ is increasing on $(a,b)$
$f'(x)<0$ on $(a,b)$	$f$ is decreasing on $(a,b)$
$f'(x_0)=0$ , sign of $f'$ changes across $x_0$	$x_0$ is a local extremum
$f'(x_0)=0$ , $f''(x_0)>0$	$x_0$ is a local minimum (graph curves upward)
$f'(x_0)=0$ , $f''(x_0)<0$	$x_0$ is a local maximum (graph curves downward)
$f''(x_0)=0$ , sign of $f''$ changes across $x_0$	$x_0$ is an inflection point

UndergraduateTwo theorems: from the Mean Value Theorem to a test for extrema

Theorem: Sufficient condition for monotonicity

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$ . If $f'(x)\ge 0$ for every $x\in(a,b)$ , and $f'(x)=0$ at only finitely many points, then $f$ is increasing on $[a,b]$ . If instead $f'(x)\le 0$ under the same condition, $f$ is decreasing on $[a,b]$ .

Why is it true?

A positive derivative means every tangent line points uphill; a curve that keeps following uphill tangents cannot come back down. Lagrange's Mean Value Theorem turns that picture into a rigorous inequality between any two points.

Proof

Take any $x_1<x_2$ in $[a,b]$ . Since $f$ is continuous on $[x_1,x_2]$ and differentiable on $(x_1,x_2)$ , Lagrange's Mean Value Theorem gives a point $c\in(x_1,x_2)$ with $f(x_2)-f(x_1)=f'(c)(x_2-x_1)$ .

Because $f'(c)\ge 0$ and $x_2-x_1>0$ , the right-hand side is $\ge 0$ , so $f(x_2)\ge f(x_1)$ . This already shows $f$ is non-decreasing on $[a,b]$ .

To upgrade "non-decreasing" to strictly "increasing" despite the finitely many zeros of $f'$ , list those zeros as $t_1<t_2<\dots<t_k$ inside $(x_1,x_2)$ . On each open piece between consecutive points of the list $\{x_1,t_1,\dots,t_k,x_2\}$ , $f'$ is strictly positive, so the same Mean Value Theorem argument applied to any two points inside that piece gives a strict inequality $f(u)<f(v)$ for $u<v$ there.

By continuity of $f$ , the strict inequalities on consecutive pieces chain together: if $x_1<u<t_1<v<x_2$ say, then $f(x_1)<f(u)<f(t_1)\le f(v)<f(x_2)$ , and since this holds for every choice of $u,v$ arbitrarily close to the endpoints, $f(x_1)<f(x_2)$ follows. Hence $f$ is (strictly) increasing on $[a,b]$ . The decreasing case follows by applying this argument to $-f$ .

Theorem: Fermat's interior extremum theorem and the second-derivative test

If $f$ is differentiable at an interior point $x_0$ of its domain and $x_0$ is a local extremum of $f$ , then $f'(x_0)=0$ . Conversely, if $f'(x_0)=0$ and $f$ is twice differentiable near $x_0$ with $f''(x_0)\ne 0$ , then $x_0$ is a local minimum when $f''(x_0)>0$ , and a local maximum when $f''(x_0)<0$ .

Why is it true?

At an extremum the tangent line must be horizontal, because the curve cannot keep rising (or falling) past a peak (or valley) without turning around; the second derivative then measures which way the curve bends at that flat point, telling a valley from a peak.

Proof

(Necessity.) Suppose $x_0$ is a local maximum; the local-minimum case is symmetric. There is $\delta>0$ with $f(x)\le f(x_0)$ for all $x$ with $|x-x_0|<\delta$ . For $h\in(0,\delta)$ , $\dfrac{f(x_0+h)-f(x_0)}{h}\le 0$ , and letting $h\to 0^+$ gives $f'(x_0)\le 0$ . For $h\in(-\delta,0)$ , the numerator is $\le 0$ and the denominator is negative, so $\dfrac{f(x_0+h)-f(x_0)}{h}\ge 0$ , and letting $h\to 0^-$ gives $f'(x_0)\ge 0$ . Since $f$ is differentiable at $x_0$ , both one-sided limits agree, forcing $f'(x_0)=0$ .

(Sufficiency, case $f''(x_0)>0$ .) Since $f'(x_0)=0$ and $f''$ exists near $x_0$ , the definition of $f''$ as the derivative of $f'$ gives $f'(x)=f'(x_0)+f''(x_0)(x-x_0)+o(x-x_0)=f''(x_0)(x-x_0)+o(x-x_0)$ as $x\to x_0$ . Because $f''(x_0)>0$ , this expression is negative for $x$ slightly less than $x_0$ and positive for $x$ slightly greater than $x_0$ .

So $f'$ changes sign from $-$ to $+$ across $x_0$ : by the monotonicity theorem, $f$ is decreasing just to the left of $x_0$ and increasing just to the right, which is exactly the definition of a local minimum at $x_0$ .

The case $f''(x_0)<0$ is identical with all inequalities reversed, giving $f'$ changing from $+$ to $-$ , hence a local maximum. When $f''(x_0)=0$ the expansion above gives no information about the sign of $f'$ near $x_0$ , so the test is inconclusive and one must examine the sign of $f'$ directly (as the pitfall below illustrates with $f(x)=x^3$ ).

AdvancedGlobal extrema on a closed interval, and asymptotes

A function continuous on a closed, bounded interval $[a,b]$ always attains a global maximum $M$ and a global minimum $m$ somewhere on $[a,b]$ (Weierstrass's extreme value theorem). Local extrema found by the sign of $f'$ are only candidates for these global values — the global maximum or minimum could instead sit at an endpoint $a$ or $b$ , where the sign-change argument does not apply. The practical algorithm is therefore: list the finitely many critical points $x_1,\dots,x_k$ in $(a,b)$ , evaluate $f$ there and at both endpoints, and compare.

M=\max_{[a,b]}f=\max\{f(a),f(x_1),\dots,f(x_k),f(b)\},\qquad m=\min_{[a,b]}f=\min\{f(a),f(x_1),\dots,f(x_k),f(b)\}

When the domain is unbounded, curves can approach a line without ever touching it — an asymptote. The line $x=x_0$ is a vertical asymptote if $\lim_{x\to x_0^\pm}f(x)=\pm\infty$ ; the line $y=L$ is a horizontal asymptote if $\lim_{x\to+\infty}f(x)=L$ or $\lim_{x\to-\infty}f(x)=L$ ; and the line $y=ax+b$ (with $a\ne 0$ ) is an oblique asymptote if $a=\lim_{x\to\infty}\dfrac{f(x)}{x}$ and $b=\lim_{x\to\infty}[f(x)-ax]$ both exist and are finite. These three cases cover every way a rational function's graph can "settle down" far from the origin.

UndergraduateReal-world applications: optimization in engineering and physics

Turning the sign of $f'$ into an algorithm is what makes derivatives a working tool rather than a curiosity: whenever a real quantity (cost, area, volume, height) can be written as a function of one variable, differentiating and applying the tests above locates the best possible choice. The two examples below come from manufacturing and from projectile motion.

Example: Maximizing the volume of an open-top box

From a square sheet of cardboard with side $12$ cm, equal squares of side $x$ are cut from each of the four corners, and the flaps are folded up to form an open-top box. Find the value of $x$ that maximizes the box's volume, and the maximum volume.

Solution

The base of the box has side $12-2x$ (each side loses two flaps of width $x$ ), and the height is $x$ , so the volume is $V(x)=x(12-2x)^2$ for $x\in(0,6)$ .

Expand or differentiate directly with the product rule: $V'(x)=(12-2x)^2+x\cdot 2(12-2x)(-2)=(12-2x)\big[(12-2x)-4x\big]=(12-2x)(12-6x)$ . Setting $V'(x)=0$ gives $x=6$ (excluded, it is an endpoint of the domain) or $x=2$ .

Since $V'(x)=6(6-x)(2-x)$ up to a positive constant near the relevant range, $V'(x)>0$ for $0<x<2$ and $V'(x)<0$ for $2<x<6$ : the sign of $V'$ changes from $+$ to $-$ at $x=2$ , so by the sign test, $x=2$ is a local (and, since it is the only critical point in $(0,6)$ , global) maximum.

The maximum volume is $V(2)=2\cdot(12-4)^2=2\cdot 64=128$ cm $^3$ .

Example: Maximum height of a thrown ball

A ball thrown straight up has height (in meters) given by $h(t)=-5t^2+20t+1$ , where $t$ is time in seconds. Find the time at which the ball reaches its maximum height, and that maximum height.

Solution

The velocity is $h'(t)=-10t+20$ . Setting $h'(t)=0$ gives $t=2$ seconds, the only critical point.

Since $h''(t)=-10<0$ for all $t$ , the second-derivative test confirms $t=2$ is a local maximum; because $h$ is a downward-opening parabola, this local maximum is also the global maximum over all $t\ge 0$ .

Substituting back, $h(2)=-5(2)^2+20(2)+1=-20+40+1=21$ . So the ball reaches its maximum height of $21$ meters at $t=2$ seconds — after that, $h'(t)<0$ and the ball is falling back down, matching the physical picture of a projectile decelerating under gravity, momentarily stopping, then descending.

Common mistake. Two classic mistakes. (1) *Assuming

f'(x_0)=0

always gives an extremum.* For

f(x)=x^3

f'(x)=3x^2

vanishes at

x_0=0

, but

f'(x)\ge 0

on both sides of

0

(it never changes sign), so

f

is increasing through

x_0=0

and

x_0

is neither a local maximum nor minimum — it is an inflection point. The second-derivative test also fails here since

f''(0)=0

too, so the sign of

f'

must be checked directly on both sides. (2) *Forgetting the endpoints when finding the global maximum/minimum on

[a,b]

.* A common error is to compute only the critical points from

f'(x)=0

and report the largest of those values as the global maximum, silently skipping

f(a)

and

f(b)

— but the true global extremum can sit exactly at an endpoint where the interior sign-change argument never applies.

Research frontier as of 2026

As of 2026, the same sign-of-the-derivative idea drives optimization at massive scale. Training a neural network is finding a (local) minimum of a loss function of millions or billions of variables; gradient descent and its variants (SGD, Adam) repeatedly move against the gradient — the multivariable analogue of $f'(x)$ — without ever computing every one of the finitely many critical points by hand. A major open challenge is understanding the geometry of these high-dimensional critical points: unlike the single-variable case where a critical point is a maximum, minimum, or inflection, in high dimensions most critical points of a random smooth function are saddle points rather than true local minima, and current research (in non-convex optimization theory) studies why gradient-based methods so often escape saddles and still find good minima in practice for deep learning loss landscapes.

On which set is $f(x)=x^3-3x$ increasing?

If $f'(x_0)=0$ and $f''(x_0)>0$ , what is $x_0$ ?

A ball thrown up has height $h(t)=-5t^2+20t+1$ meters. At what time $t$ does it reach maximum height?

Cutting squares of side $x$ from a $12$ cm square sheet and folding up the sides forms an open-top box. Which $x$ maximizes the volume?

References

Michael Spivak (2008). Calculus
Stephen Boyd, Lieven Vandenberghe (2004). Convex Optimization