Chapter 04

Medium

Applications of Derivatives — Core

00 · Symbol Glossary

$f'(x) > 0$f prime positive — increasing

The condition that the derivative is positive on an interval, meaning the function's outputs are rising as $x$ increases across that interval.

$f'(c) = 0 \text{ or DNE}$critical point condition

The defining condition for a critical point at $x=c$ : either the derivative is exactly zero there, or the derivative fails to exist there (e.g. a corner, as with $\lvert x\rvert$ in Chapter 02).

$f''(x)$f double prime — concavity indicator

Introduced in Chapter 02 as the second derivative. In this chapter its sign (not just its value) becomes the tool for describing how a curve bends.

$\text{c.p.}$critical point (abbreviation)

Shorthand used in step-by-step work for "critical point," to avoid repeating the full phrase inside computations.

01 · Increasing and Decreasing Functions

The sign of the derivative reveals the direction a function is moving — not by looking at the graph, but by computing $f'(x)$ and checking whether it is positive or negative.

Definition — Increasing and Decreasing on an Interval

$f$ is increasing on an interval if $f'(x) > 0$ for every $x$ in that interval.

$f$ is decreasing on an interval if $f'(x) < 0$ for every $x$ in that interval.

$f$ is (momentarily) stationary where $f'(x) = 0$ .

Step-by-step — Find where $f(x) = x^3 - 3x^2 - 9x + 5$ is increasing/decreasing

Differentiate:

f'(x) = 3x^2 - 6x - 9

(power rule term by term, Chapter 02).

Set $f'(x)=0$ to find candidate sign-change points:

3x^2-6x-9=0

. Divide every term by

3

x^2-2x-3=0

Factor:

x^2-2x-3 = (x-3)(x+1)

, since

-3\times1=-3

and

-3+1=-2

match the coefficients. Roots:

x=3

and

x=-1

Test a value in each of the three intervals formed by $-1$ and $3$ : at

x=-2

(left of

-1

f'(-2)=3(4)-6(-2)-9=12+12-9=15>0

— increasing. At

x=0

(between

-1

and

3

f'(0)=0-0-9=-9<0

— decreasing. At

x=4

(right of

3

f'(4)=3(16)-6(4)-9=48-24-9=15>0

— increasing.

Conclusion:

f

is increasing on

(-\infty,-1)

, decreasing on

(-1,3)

, and increasing again on

(3,\infty)

02 · Critical Points and the First Derivative Test

The points where a function switches from increasing to decreasing (or vice versa) are exactly the boundaries found in Section 01 — these are called critical points, and they are where local peaks and valleys occur.

Definition — Critical Point

$x=c$ is a critical point of $f$ if $c$ is in the domain of $f$ and either $f'(c)=0$ or $f'(c)$ does not exist.

Definition — First Derivative Test

At a critical point $x=c$ :

If $f'$ changes from positive to negative at $c$ , $f$ has a local maximum at $c$ .
If $f'$ changes from negative to positive at $c$ , $f$ has a local minimum at $c$ .
If $f'$ does not change sign at $c$ , $f$ has neither — the curve flattens momentarily but keeps the same direction.

Example — Classifying critical points from Section 01

For $f(x)=x^3-3x^2-9x+5$ , the critical points found above are $x=-1$ and $x=3$ . From the sign pattern (increasing, then decreasing, then increasing): at $x=-1$ , $f'$ goes from $+$ to $-$ — a local maximum. At $x=3$ , $f'$ goes from $-$ to $+$ — a local minimum.

Compute the values: $f(-1) = (-1)^3-3(-1)^2-9(-1)+5 = -1-3+9+5=10$ . $f(3)=27-27-27+5=-22$ . Local max of $10$ at $x=-1$ ; local min of $-22$ at $x=3$ .

❌ Assuming every critical point is a local extremum

For $f(x)=x^3$ , $f'(x)=3x^2$ , so $f'(0)=0$ — $x=0$ is a critical point.

Check the sign of $f'$ on either side: at $x=-1$ , $f'(-1)=3(1)=3>0$ . At $x=1$ , $f'(1)=3(1)=3>0$ . The derivative is positive on both sides — it never changes sign.

Why it breaks: the first derivative test requires a sign change to classify a critical point as a max or min. Here $f'$ touches $0$ at $x=0$ but stays non-negative throughout, so $x^3$ is increasing right through that point.

Consequence: $x=0$ is a critical point of $f(x)=x^3$ but is neither a local max nor a local min — it is an inflection point (Section 04), where the curve flattens momentarily but does not turn around.

03 · Concavity and the Second Derivative

Beyond direction, a curve also has a shape — it can bend upward like a cup or downward like a dome. The second derivative, $f''(x)$ , measures this bending.

Definition — Concavity

$f$ is concave up on an interval if $f''(x) > 0$ there — the curve bends like a cup, and the tangent lines lie below the curve.

$f$ is concave down on an interval if $f''(x) < 0$ there — the curve bends like a dome, and the tangent lines lie above the curve.

Example — Concavity of $f(x)=x^2$

$f'(x)=2x$ , $f''(x)=2$ . Since $f''(x)=2>0$ everywhere, $f(x)=x^2$ is concave up on its entire domain — consistent with its familiar cup shape.

04 · Inflection Points and the Second Derivative Test

Definition — Inflection Point

An inflection point occurs at $x=c$ where the concavity changes — $f''(x)$ switches sign as $x$ passes through $c$ (from positive to negative, or negative to positive).

Definition — Second Derivative Test

At a critical point $x=c$ where $f'(c)=0$ :

If $f''(c) > 0$ , $f$ has a local minimum at $c$ (concave up — curve opens upward around the flat point).
If $f''(c) < 0$ , $f$ has a local maximum at $c$ (concave down — curve opens downward around the flat point).
If $f''(c) = 0$ , the test is inconclusive — fall back to the first derivative test.

Step-by-step — Apply the second derivative test to $f(x)=x^3-3x^2-9x+5$

Recall the critical points and second derivative:

x=-1

and

x=3

(from Section 01–02);

f'(x)=3x^2-6x-9

, so

f''(x) = 6x-6

Evaluate $f''$ at $x=-1$ :

f''(-1) = 6(-1)-6 = -6-6=-12<0

— concave down, so local maximum, matching Section 02's first-derivative-test conclusion.

Evaluate $f''$ at $x=3$ :

f''(3) = 6(3)-6=18-6=12>0

— concave up, so local minimum, again matching Section 02.

Find the inflection point by setting $f''(x)=0$ :

6x-6=0 \implies x=1

. Check the sign change:

f''(0)=6(0)-6=-6<0

;

f''(2)=6(2)-6=6>0

. Sign changes from negative to positive at

x=1

— confirmed inflection point.

❌ f''(c) = 0 does not automatically mean an inflection point

For $f(x)=x^4$ , $f'(x)=4x^3$ , $f''(x)=12x^2$ . Setting $f''(x)=0$ gives $x=0$ .

Why it breaks: an inflection point requires concavity to actually change sign, not merely touch zero. Check both sides of $x=0$ : $f''(-1)=12(1)=12>0$ and $f''(1)=12(1)=12>0$ — concave up on both sides.

Consequence: $x=0$ is not an inflection point of $x^4$ , even though $f''(0)=0$ — the concavity never flips. (It is, however, a local minimum by the first derivative test: $f'(x)=4x^3$ goes from negative to positive at $x=0$ .)

05 · Curve Sketching

Combining Sections 01–04 gives a full qualitative picture of a function's graph without needing to plot individual points.

Step-by-step — Sketch-level analysis of $f(x)=x^3-3x^2-9x+5$

Increasing/decreasing (Section 01): increasing on

(-\infty,-1)

, decreasing on

(-1,3)

, increasing on

(3,\infty)

Local extrema (Section 02): local max

10

x=-1

; local min

-22

x=3

Concavity (Section 03–04): concave down on

(-\infty,1)

, concave up on

(1,\infty)

; inflection point at

x=1

, where

f(1)=1-3-9+5=-6

Assemble the description: the curve rises to a peak at

(-1,10)

, falls through an inflection at

(1,-6)

while still descending, bottoms out at

(3,-22)

, then rises indefinitely — a complete shape description using only algebra, no plotting required.

06 · L'Hôpital's Rule

Chapter 01 left the indeterminate form $\frac{0}{0}$ to be resolved by algebra (factoring). Derivatives supply a second, often faster tool for that same problem.

Definition — L'Hôpital's Rule

If $\lim_{x\to a}\dfrac{f(x)}{g(x)}$ produces the indeterminate form $\dfrac{0}{0}$ or $\dfrac{\infty}{\infty}$ , and $f,g$ are differentiable near $a$ with $g'(x)\neq0$ near $a$ , then

\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}

provided the right-hand limit exists. This differentiates the top and bottom separately — it is not the quotient rule, and the denominator is never squared.

Step-by-step — Evaluate $\lim_{x\to0}\dfrac{\sin x}{x}$-style limit $\lim_{x\to2}\dfrac{x^2-4}{x-2}$ via L'Hôpital

Check the indeterminate form: substituting

x=2

gives

\dfrac{4-4}{0}=\dfrac{0}{0}

— L'Hôpital applies.

Differentiate numerator and denominator separately:

\dfrac{d}{dx}[x^2-4]=2x

;

\dfrac{d}{dx}[x-2]=1

Form the new limit:

\lim_{x\to2}\dfrac{2x}{1}

Substitute $x=2$ :

2(2)=4

. So

\lim_{x\to2}\dfrac{x^2-4}{x-2}=4

— matching the factoring method:

\dfrac{(x-2)(x+2)}{x-2}=x+2\to4

x\to2

. ✓

❌ Applying L'Hôpital's rule when the form is not indeterminate

For $\lim_{x\to1}\dfrac{x^2+3}{x+1}$ , substituting gives $\dfrac{1+3}{1+1}=\dfrac{4}{2}=2$ — a perfectly defined value, not $\frac00$ or $\frac{\infty}{\infty}$ .

Why it breaks: applying L'Hôpital anyway — differentiating to get $\lim_{x\to1}\dfrac{2x}{1}=2$ — happens to land on the same numeric answer here by coincidence, but the method itself is invalid whenever the form is not indeterminate to begin with.

Consequence: always verify the $\frac00$ or $\frac{\infty}{\infty}$ form before differentiating top and bottom. Applying L'Hôpital to a non-indeterminate quotient is not justified by the theorem and can give a wrong answer on other examples where the direct value and the differentiated value do not coincide.

07 · Exercises

EXERCISE 4.1

Differentiate, set the derivative to zero, factor, and test the sign of $f'$ in each resulting interval.

$f(x)=x^3-12x$ . $f'(x)=3x^2-12=3(x^2-4)=3(x-2)(x+2)$ . Zero at $x=\pm2$ . Test $x=-3$ : $f'(-3)=3(9-4)=15>0$ . Test $x=0$ : $f'(0)=3(-4)=-12<0$ . Test $x=3$ : $f'(3)=3(5)=15>0$ . Increasing on $(-\infty,-2)$ , decreasing on $(-2,2)$ , increasing on $(2,\infty)$ .

Find the intervals of increase and decrease for $f(x) = x^3-12x$ .

EXERCISE 4.2

Use the sign pattern from Exercise 4.1 to classify each critical point with the first derivative test.

From 4.1, $f'$ goes $+,-,+$ around $x=-2,2$ . At $x=-2$ : $+$ to $-$ , local max. $f(-2)=-8+24=16$ . At $x=2$ : $-$ to $+$ , local min. $f(2)=8-24=-16$ .

Using the results of Exercise 4.1, classify the critical points of $f(x)=x^3-12x$ as local maxima or minima and give the function values there.

EXERCISE 4.3

Compute $f''(x)$ , set it to zero, and check that the sign actually changes across that point.

$f(x)=x^3-12x$ , $f'(x)=3x^2-12$ , $f''(x)=6x$ . Zero at $x=0$ . $f''(-1)=-6<0$ ; $f''(1)=6>0$ — sign changes, confirmed inflection point at $x=0$ , where $f(0)=0$ .

Find the inflection point of $f(x)=x^3-12x$ , verifying the concavity actually changes sign.

EXERCISE 4.4

Apply the second derivative test using $f''(x)$ evaluated at the critical points from Exercise 4.1.

$f''(x)=6x$ . At $x=-2$ : $f''(-2)=-12<0$ , concave down, local max (matches 4.2). At $x=2$ : $f''(2)=12>0$ , concave up, local min (matches 4.2).

Verify the classifications from Exercise 4.2 using the second derivative test instead.

EXERCISE 4.5

Construct a function whose second derivative touches zero without changing sign — mirror the $x^4$ example in Section 04.

$f(x)=x^4-4x^3+6x^2$ works: $f'(x)=4x^3-12x^2+12x$ , $f''(x)=12x^2-24x+12=12(x-1)^2$ . This is $\geq0$ everywhere and equals $0$ only at $x=1$ , never going negative. So $f''$ touches $0$ at $x=1$ but never changes sign — $x=1$ is not an inflection point despite $f''(1)=0$ .

Give an example of a function where $f''(c)=0$ at some point $c$ , but $c$ is not an inflection point. Justify by checking the sign of $f''$ on both sides.

EXERCISE 4.6

Check whether direct substitution gives an indeterminate form before differentiating top and bottom separately.

$\lim_{x\to0}\dfrac{e^{2x}-1}{x}$ ... using instead a polynomial form to stay within tools introduced so far: $\lim_{x\to3}\dfrac{x^2-9}{x^2-3x}$ . Substituting $x=3$ : $\dfrac{0}{0}$ — indeterminate, L'Hôpital applies. Differentiate: numerator $\to2x$ , denominator $\to2x-3$ . New limit: $\lim_{x\to3}\dfrac{2x}{2x-3}=\dfrac{6}{3}=2$ . Check by factoring: $\dfrac{(x-3)(x+3)}{x(x-3)}=\dfrac{x+3}{x}\to\dfrac{6}{3}=2$ as $x\to3$ . ✓ Matches.

Evaluate $\lim_{x\to3}\dfrac{x^2-9}{x^2-3x}$ using L'Hôpital's rule, and verify the result by factoring instead.

08 · Chapter Summary

Concept	Formula / Rule
Increasing / decreasing	$f'(x)>0$ / $f'(x)<0$ on an interval
Critical point	$f'(c)=0$ or $f'(c)$ undefined
First derivative test	Sign change of $f'$ at $c$ classifies local max/min
Concave up / down	$f''(x)>0$ / $f''(x)<0$
Inflection point	$f''(x)$ changes sign at $c$
Second derivative test	$f''(c)>0\Rightarrow$ local min; $f''(c)<0\Rightarrow$ local max; $=0\Rightarrow$ inconclusive
L'Hôpital's rule	$\lim\frac{f}{g}=\lim\frac{f'}{g'}$ for $\frac00$ or $\frac{\infty}{\infty}$ forms only

Next: Chapter 05 — Applications of Derivatives: Optimization, where increasing/decreasing and concavity analysis are used to find the single best value of a quantity, not just its local behavior.