Unit 5: Analytical Applications of Differentiation

Mean Value Theorem (MVT) and Rolle’s Theorem

Theorem (Mean Value Theorem). If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that

$$f'(c) = \frac{f(b)-f(a)}{b-a}.$$

Rolle’s Theorem is the special case where $f(a)=f(b)$.

The Mean Value Theorem says that under the right smoothness conditions, some instantaneous rate equals the average rate over the interval.

The hypotheses are essential:

Conditions

• continuous on $[a,b]$,
• differentiable on $(a,b)$.

If either hypothesis fails, the theorem may not apply, even if the conclusion happens to be true.

Rolle’s Theorem is the same idea when the average rate is zero. If a smooth function starts and ends at the same height, then somewhere in between it has a horizontal tangent.

Proof (MVT). Define the secant-line function through the endpoints:

$$L(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a).$$

Now define

$$g(x)=f(x)-L(x).$$

Because $f$ and $L$ are continuous on $[a,b]$ and differentiable on $(a,b)$, so is $g$. Also,

$$g(a)=f(a)-L(a)=0,$$

and

$$g(b)=f(b)-L(b)=0.$$

By Rolle’s Theorem, there is some $c\in(a,b)$ such that $g'(c)=0$. Since

$$g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a},$$

we get

$$0=f'(c)-\frac{f(b)-f(a)}{b-a}.$$

Therefore,

$$f'(c)=\frac{f(b)-f(a)}{b-a}.$$

Example. Verify Rolle’s Theorem for

$$f(x)=x^2-4x+3$$

on $[1,3]$, then find the value of $c$.

The function is a polynomial, so it is continuous on $[1,3]$ and differentiable on $(1,3)$. Check the endpoint values:

$$f(1)=1-4+3=0,$$

and

$$f(3)=9-12+3=0.$$

Since $f(1)=f(3)$, Rolle’s Theorem applies. Differentiate:

$$f'(x)=2x-4.$$

Set $f'(c)=0$:

$$2c-4=0 \quad\Longrightarrow\quad c=2.$$

The value $c=2$ lies in $(1,3)$, so it is the point guaranteed by Rolle’s Theorem.

Example. For $f(x)=x^2$ on $[1,3]$, find the value $c$ guaranteed by the Mean Value Theorem.

The function is a polynomial, so it is continuous on $[1,3]$ and differentiable on $(1,3)$, and the theorem applies. The average rate of change is

$$\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}=4.$$

Since $f'(x)=2x$, set the instantaneous rate equal to the average rate:

$$2c=4\quad\Longrightarrow\quad c=2.$$

The value $c=2$ lies in $(1,3)$, so it is the value guaranteed by the theorem.

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Extreme Value Theorem (EVT) and critical points

The Extreme Value Theorem (EVT) guarantees that a continuous function on a closed interval has both an absolute maximum and an absolute minimum.

Theorem (EVT). If $f$ is continuous on $[a,b]$, then there are numbers $m$ and $M$ in $[a,b]$ such that

$$f(m)\le f(x)\le f(M)$$

for every $x$ in $[a,b]$.

The proof for EVT is based in non-calculus fields like real analysis so it will not be shown here.

The two conditions matter. The interval must be closed, so endpoints are included, and the function must be continuous, so it cannot jump over or miss its highest or lowest value. EVT tells you the extrema exist; it does not tell you where they are. Critical points and endpoints are how you find the candidates.

Definition. A critical point of $f$ occurs at $x=c$ when:

• $f'(c) = 0$, or
• $f'(c)$ does not exist,

provided $c$ is in the domain of $f$.

If $f$ has a local extremum at an interior point $c$ and $f'(c)$ exists, then

$$f'(c)=0.$$

This theorem explains why critical numbers matter, but it does not say every critical number is an extremum. It only says that if an interior local extremum happens at a differentiable point, the derivative must be zero there. A sign test or value comparison is still needed.

There are two different candidate lists:

• for local extrema, check interior critical points,
• for absolute extrema on a closed interval, check interior critical points and endpoints.

Endpoints matter for absolute extrema because EVT guarantees the largest and smallest output somewhere on the whole closed interval. Endpoints are included in that interval, even though they are not usually called local extrema in AP Calculus.

When justifying an absolute maximum or minimum, compare function values, not derivative values. The derivative helps you find candidates, but the output values decide the final answer.

Problem-solving strategy

1. Check that $f$ is continuous on the closed interval.
2. Find critical points inside the interval.
3. Evaluate $f$ at those critical points and at both endpoints.
4. Compare the output values.

Warning

A point where $f'(c)=0$ is not automatically a maximum or minimum. For example, $f(x)=x^3$ has $f'(0)=0$, but $x=0$ is not an extremum.

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Graphical meaning of derivatives

When looking at a function, its first derivative can tell you a lot about the direction:

• $f'(x) > 0$ on an interval implies $f$ is increasing there.
• $f'(x) < 0$ on an interval implies $f$ is decreasing there.

Sign charts are the cleanest way to justify interval behavior.

When looking at a function, its second derivative can tell you a lot about the shape:

• $f''(x)>0$, then $f'(x)$ is increasing. The graph of $f$ bends upward (concave up) because its slopes are becoming more positive or less negative.
• $f''(x)<0$, then $f'(x)$ is decreasing. The graph of $f$ bends downward (concave down) because its slopes are becoming less positive or more negative.

An inflection point is a point where concavity changes. The equation $f''(x)=0$ only gives a possible location; you still need to check that the concavity actually changes.

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The First Derivative Test

A very useful test to find local extrema is the First Derivative Test. At points where $f'(x)=0$ (critical points), we can use this test to see if there is a local minimum, local maximum, or neither.

Theorem (First Derivative Test). If $f'$ changes:

• positive to negative at $c$: local maximum,
• negative to positive at $c$: local minimum,
• no sign change: neither.

Proof (First Derivative Test). Suppose $f'$ changes from positive to negative at $c$. Then $f$ is increasing just to the left of $c$ and decreasing just to the right of $c$. So values near $c$ on the left rise toward $f(c)$, and values near $c$ on the right fall away from $f(c)$. This makes $f(c)$ a local maximum.

If $f'$ changes from negative to positive, the same reasoning reverses: $f$ decreases into $c$ and increases after $c$, so $f(c)$ is a local minimum.

If $f'$ does not change sign, then $f$ keeps increasing on both sides or keeps decreasing on both sides. In that case, $c$ is not a local extremum.

Example. Find and classify the critical points of $f(x)=x^3-3x^2$.

Differentiate and factor:

$$f'(x)=3x^2-6x=3x(x-2).$$

So $f'(x)=0$ at $x=0$ and $x=2$. Test the sign of $f'$ on each interval:

Test one point in each interval:

$$f'(-1)=3(-1)(-3)=9>0.$$ $$f'(1)=3(1)(-1)=-3<0.$$ $$f'(3)=3(3)(1)=9>0.$$

The derivative goes positive to negative at $x=0$, so $f$ has a local maximum there, with $f(0)=0$. The derivative goes negative to positive at $x=2$, so $f$ has a local minimum there, with $f(2)=8-12=-4$.

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Absolute extrema on a closed interval

Definition. The absolute extrema of an interval $[a,b]$ is defined as the point $x=c$ where

• $f(c)$ is at a maximum (for absolute maxima),
• $f(c)$ is at a minimum (for absolute minima).

To find absolute max/min (extrema) of $f$ on $[a,b]$:

Working checklist

1. Find critical points inside $(a,b)$.
2. Evaluate $f$ at each critical point.
3. Evaluate $f(a)$ and $f(b)$.
4. Compare all values to find the extrema.

The largest output is the absolute maximum, and the smallest output is the absolute minimum. Endpoints must be included because absolute extrema can occur at endpoints even though local extrema usually focus on interior points.

Example. Find the absolute maximum and minimum of $f(x)=x^3-3x$ on $[0,2]$.

Differentiate to locate interior critical points:

$$f'(x)=3x^2-3=3(x^2-1),$$

so $f'(x)=0$ at $x=\pm1$. Only $x=1$ lies inside $(0,2)$. Evaluate $f$ at this critical point and at both endpoints:

$$f(0)=0,\qquad f(1)=1-3=-2,\qquad f(2)=8-6=2.$$

Comparing the candidate values, the absolute maximum is $2$ at $x=2$, and the absolute minimum is $-2$ at $x=1$.

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Concavity and the second derivative

Concavity describes how the slopes of a function are changing.

• If $f''(x)>0$, then $f'$ is increasing and the graph is concave up.
• If $f''(x)<0$, then $f'$ is decreasing and the graph is concave down.

An inflection point occurs where concavity changes. The equation $f''(x)=0$ gives possible inflection points, but it is not enough by itself. You still need to check that $f''$ changes sign or that the graph actually changes concavity.

Example. Find the concavity intervals and inflection point of

$$f(x)=x^3-3x^2.$$

Compute the second derivative:

$$f'(x)=3x^2-6x, \qquad f''(x)=6x-6.$$

Set $f''(x)=0$:

$$6x-6=0 \quad\Longrightarrow\quad x=1.$$

Test around $x=1$. If $x<1$, then $f''(x)<0$, so the graph is concave down. If $x>1$, then $f''(x)>0$, so the graph is concave up. Since concavity changes at $x=1$, there is an inflection point there:

$$f(1)=1-3=-2.$$

The inflection point is $(1,-2)$.

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The Second Derivative Test

The Second Derivative Test is an alternate way of finding minima and maxima.

Theorem (Second Derivative Test). If $f'(c)=0$ and:

• $f''(c)>0$, then $f$ has a local minimum at $c$,
• $f''(c)<0$, then $f$ has a local maximum at $c$,
• $f''(c)=0$, the test is inconclusive.

Example. Use the second derivative test to classify the critical points of $f(x)=x^3-12x$.

First find the critical points:

$$f'(x)=3x^2-12=3(x^2-4)=3(x-2)(x+2),$$

so $f'(x)=0$ at $x=-2$ and $x=2$. The second derivative is

$$f''(x)=6x.$$

Evaluate at each critical point:

$$f''(2)=12>0,\qquad f''(-2)=-12<0.$$

Since $f''(2)>0$, there is a local minimum at $x=2$, with $f(2)=8-24=-16$. Since $f''(-2)<0$, there is a local maximum at $x=-2$, with $f(-2)=-8+24=16$.

Choosing between the tests

The First Derivative Test is usually more reliable because it checks what the function is actually doing on both sides of the critical point. It works for critical points where $f'(c)=0$ and for critical points where $f'(c)$ does not exist, as long as you can test signs around $c$.

The Second Derivative Test is faster when it applies, but it has two limits:

• it only applies when $f'(c)=0$,
• if $f''(c)=0$, the test gives no conclusion.

Tip

Tip

Use the First Derivative Test when a sign chart is already natural or when $f'$ is undefined at the critical point. Use the Second Derivative Test when $f''(c)$ is easy to compute and clearly positive or negative.

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Curve sketching framework

A solid derivative-based sketch includes:

• intercepts,
• asymptotes if relevant,
• intervals increasing/decreasing,
• local extrema,
• intervals concave up/down,
• inflection points,
• end behavior.

You do not need a perfect drawing at first. The goal is to collect enough structure that the graph has the correct shape. Start with the domain and discontinuities, then use derivatives to decide how the graph moves.

Working checklist

1. Find the domain and any vertical asymptotes or holes.
2. Find intercepts if they are easy.
3. Use $f'$ to make increasing/decreasing intervals.
4. Use $f''$ to make concavity intervals.
5. Mark local extrema and inflection points.
6. Add end behavior or horizontal/slant asymptotes.

Warning

A critical point is only a candidate. Do not label it a maximum or minimum until a derivative sign change, second derivative test, or value comparison supports it.

Example. Sketch the important derivative information for

$$f(x)=x^3-3x^2.$$

First derivative:

$$f'(x)=3x^2-6x=3x(x-2).$$

The critical points are $x=0$ and $x=2$. A sign chart for $f'$ gives:

• increasing on $(-\infty,0)$,
• decreasing on $(0,2)$,
• increasing on $(2,\infty)$.

So $x=0$ is a local maximum and $x=2$ is a local minimum.

Second derivative:

$$f''(x)=6x-6.$$

The graph is concave down on $(-\infty,1)$ and concave up on $(1,\infty)$, so there is an inflection point at $x=1$.

Evaluate the key points:

$$f(0)=0,\qquad f(1)=-2,\qquad f(2)=-4.$$

A good sketch should pass through those points, rise before $0$, fall between $0$ and $2$, rise after $2$, and change concavity at $x=1$.

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Optimization

Optimization problems mix modeling with calculus. The derivative only works after the quantity being optimized is written as a one-variable function. Most optimization problems ask for a local or absolute minimum (such as the least time) or maximum (such as the greatest profit).

The hardest part of optimization is usually building the one-variable function. A good setup keeps three pieces separate:

• the target quantity, which is what you want to maximize or minimize,
• the constraint equation, which connects the variables,
• the feasible domain, which says what values make sense in the context.

If the target quantity has two variables, use the constraint to solve for one variable and substitute. After that, the calculus part is standard: differentiate, find critical points, and test candidates.

Optimization problems often fail because the domain is ignored. After writing the target function, determine the allowed interval from the context:

• lengths must usually be positive,
• time may be restricted to a stated interval,
• square roots require nonnegative radicands,
• denominators cannot be zero.

If the allowed domain is closed, use endpoint comparison. If the domain is open or unbounded, use derivative signs and behavior at the ends.

Problem-solving strategy

1. Identify the quantity to optimize.
2. Write it as a function of one variable.
3. Determine the feasible domain.
4. Differentiate and find critical points.
5. Test candidates and interpret.

Common mistakes

• Optimizing the constraint instead of the target quantity.
• Forgetting endpoints on a closed interval.
• Ignoring domain restrictions like positive lengths or nonzero denominators.
• Reporting only the critical value of the variable instead of answering the original question.

Example. A rectangular pen is built with $40$ meters of fencing. What dimensions maximize the enclosed area?

Let the rectangle have width $x$ and height $y$. The perimeter constraint is

$$2x+2y=40\quad\Longrightarrow\quad y=20-x.$$

The area, written as a one-variable function, is

$$A(x)=x(20-x)=20x-x^2,\qquad 0<x<20.$$

Differentiate and find the critical point:

$$A'(x)=20-2x=0\quad\Longrightarrow\quad x=10.$$

Since $A''(x)=-2<0$, this critical point is a maximum. Then $y=20-10=10$, so the pen is a $10\times10$ square with maximum area

$$A(10)=10\cdot10=100\ \text{square meters}.$$

Example. A rectangular box with a square base and no top must have volume $500$ cubic centimeters. Find the dimensions that minimize the surface area.

Let $x$ be the side length of the square base and $h$ be the height. The volume constraint is

$$x^2h=500,$$

so

$$h=\frac{500}{x^2}.$$

The surface area includes the square base and four side rectangles:

$$S=x^2+4xh.$$

Substitute the constraint:

$$S(x)=x^2+4x\left(\frac{500}{x^2}\right)=x^2+\frac{2000}{x}, \qquad x>0.$$

Differentiate:

$$S'(x)=2x-\frac{2000}{x^2}.$$

Set the derivative equal to zero:

$$2x-\frac{2000}{x^2}=0.$$

Multiply by $x^2$:

$$2x^3-2000=0 \quad\Longrightarrow\quad x^3=1000 \quad\Longrightarrow\quad x=10.$$

Then

$$h=\frac{500}{10^2}=5.$$

Because $S(x)\to\infty$ as $x\to0^+$ and as $x\to\infty$, the critical point gives the absolute minimum. The box should have square base side length $10$ cm and height $5$ cm.

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Newton’s method

As a reminder, linearization is the process of estimating the value of a function using its derivative:

$$L(x) = f(a)+f'(a)(x-a).$$

Using linearization, we can approximate the roots of a function using Newton’s method (also known as the Newton-Raphson method) for approximating roots. Newton’s method uses tangent lines to approximate roots.

Theorem (Newton’s Method). To approximate a root of $f(x)=0$, start with a guess $x_n$ and repeatedly use

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.$$

$x_n$ represents the $n$th iteration, where you keep using your previous attempts to estimate the root. The first number always starts out as a guess, as after many iterations, the value will get closer and closer to the real root. However, the method works best when the starting guess is close to the root and the derivative is not near zero. If the tangent line is nearly horizontal, the next approximation can jump far away.

Proof (Newton’s Method). The linearization of $f$ at $x_n$ is

$$L(x)=f(x_n)+f'(x_n)(x-x_n).$$

Newton’s method uses the root of this tangent line as the next guess, so set $L(x)=0$:

$$0=f(x_n)+f'(x_n)(x-x_n).$$

Solve for $x$:

$$f'(x_n)(x-x_n)=-f(x_n),$$ $$x-x_n=-\frac{f(x_n)}{f'(x_n)}.$$

Therefore,

$$x=x_n-\frac{f(x_n)}{f'(x_n)}.$$

This new $x$-value is called $x_{n+1}$.

Example. Use Newton’s method to approximate $\sqrt2$ by solving $x^2-2=0$, starting with $x_0=1.5$.

Let

$$f(x)=x^2-2, \qquad f'(x)=2x.$$

The Newton update is

$$x_{n+1}=x_n-\frac{x_n^2-2}{2x_n}.$$

Starting with $x_0=1.5$:

$$x_1=1.5-\frac{1.5^2-2}{2(1.5)} =1.5-\frac{0.25}{3} \approx1.4167.$$

Apply the formula again:

$$x_2=1.4167-\frac{1.4167^2-2}{2(1.4167)} \approx1.4142.$$

So $\sqrt2\approx1.4142$.

Example. Use Newton’s method to approximate a solution of $\cos x=x$, starting with $x_0=1$.

Write the equation as a root problem:

$$f(x)=\cos x-x.$$

Then

$$f'(x)=-\sin x-1.$$

Newton’s method gives

$$x_{n+1}=x_n-\frac{\cos x_n-x_n}{-\sin x_n-1}.$$

Starting with $x_0=1$:

$$x_1=1-\frac{\cos(1)-1}{-\sin(1)-1}\approx0.7504.$$

One more iteration gives

$$x_2=0.7504-\frac{\cos(0.7504)-0.7504}{-\sin(0.7504)-1}\approx0.7391.$$

So the solution is approximately $x\approx0.7391$.

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Tips for the exam

Analytical applications ask you to turn derivative information into a story about the original function. The safest approach is to separate where candidates occur from what those candidates mean.

Problem-solving strategy

1. Find the domain first, because endpoints and discontinuities matter.
2. Find critical numbers from $f'(x)=0$ and places where $f'$ is undefined.
3. Build a sign chart for $f'$ to determine increasing/decreasing behavior.
4. Build a sign chart for $f''$ to determine concavity.
5. Use endpoint values and critical values for absolute extrema on closed intervals.
6. Use derivative sign changes, not just derivative zeros, to justify local extrema.

On the examJustification language

Write conclusions in terms of signs and changes: ”$f$ has a local maximum at $x=c$ because $f'$ changes from positive to negative there.” This is stronger than saying ”$f'(c)=0$.”

For optimization, the derivative work is only the middle of the problem. A complete solution should also define variables, state the constraint, state the domain, and answer in the original units.