Chapter 4: Applications of Derivatives

Example 1: Expanding Circle

A stone is dropped into a still pond, producing a circular ripple whose radius increases at a constant rate of $3$ ft/s. How fast is the area of the circle increasing when the radius is $10$ ft?

Solution. Let $r$ be the radius and $A$ the area of the circle at time $t$. We are given $\frac{dr}{dt} = 3$ ft/s and want $\frac{dA}{dt}$ when $r = 10$ ft.

The relationship between area and radius is:

$$A = \pi r^2$$

Differentiate both sides with respect to $t$:

$$\frac{dA}{dt} = 2\pi r \, \frac{dr}{dt}$$

Substitute $r = 10$ and $\frac{dr}{dt} = 3$:

$$\frac{dA}{dt} = 2\pi(10)(3) = 60\pi \approx 188.5 \text{ ft}^2\text{/s}$$

Example 2: Sliding Ladder

A 13-foot ladder leans against a vertical wall. The base of the ladder slides away from the wall at $2$ ft/s. How fast is the top of the ladder sliding down the wall when the base is $5$ ft from the wall?

Solution. Let $x$ be the distance from the base of the ladder to the wall, and let $y$ be the height of the top of the ladder on the wall. The ladder length is constant at 13 ft, so by the Pythagorean theorem:

$$x^2 + y^2 = 169$$

Differentiate with respect to $t$:

$$2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} = 0$$

When $x = 5$, we find $y = \sqrt{169 - 25} = \sqrt{144} = 12$. We are given $\frac{dx}{dt} = 2$ ft/s. Substitute:

$$2(5)(2) + 2(12)\frac{dy}{dt} = 0$$ $$20 + 24\frac{dy}{dt} = 0$$ $$\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6} \text{ ft/s}$$

The negative sign confirms the top is sliding down at $\frac{5}{6}$ ft/s.

Example 3: Filling a Conical Tank

Water flows into an inverted conical tank at a rate of $8$ ft$^3$/min. The tank has a radius of $4$ ft at the top and a height of $10$ ft. How fast is the water level rising when the water is $5$ ft deep?

Solution. Let $r$ be the radius of the water surface and $h$ the depth of water at time $t$. We know $\frac{dV}{dt} = 8$ ft$^3$/min and want $\frac{dh}{dt}$ when $h = 5$ ft.

The volume of a cone is $V = \frac{1}{3}\pi r^2 h$. By similar triangles with the tank dimensions, $\frac{r}{h} = \frac{4}{10} = \frac{2}{5}$, so $r = \frac{2h}{5}$. Substituting to eliminate $r$:

$$V = \frac{1}{3}\pi \left(\frac{2h}{5}\right)^2 h = \frac{1}{3}\pi \cdot \frac{4h^2}{25} \cdot h = \frac{4\pi h^3}{75}$$

Differentiate with respect to $t$:

$$\frac{dV}{dt} = \frac{4\pi}{75} \cdot 3h^2 \, \frac{dh}{dt} = \frac{4\pi h^2}{25}\,\frac{dh}{dt}$$

Substitute $\frac{dV}{dt} = 8$ and $h = 5$:

$$8 = \frac{4\pi(25)}{25}\,\frac{dh}{dt} = 4\pi\,\frac{dh}{dt}$$ $$\frac{dh}{dt} = \frac{8}{4\pi} = \frac{2}{\pi} \approx 0.637 \text{ ft/min}$$

Example 4: Expanding Sphere

A spherical balloon is being inflated so that its radius increases at a constant rate of $2$ cm/s. At the instant when $r = 5$ cm:

1. How fast is the volume increasing?
2. How fast is the surface area increasing?

Setup. Let $r = r(t)$ be the radius, $V = \frac{4}{3}\pi r^3$ the volume, and $S = 4\pi r^2$ the surface area. We are given $\dfrac{dr}{dt} = 2$ cm/s.

Part (a) — Volume. Differentiate $V = \frac{4}{3}\pi r^3$ with respect to $t$:

$$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$$

Substitute $r = 5$ and $\dfrac{dr}{dt} = 2$:

$$\frac{dV}{dt} = 4\pi(25)(2) = 200\pi \approx 628.3 \text{ cm}^3/\text{s}$$

Part (b) — Surface Area. Differentiate $S = 4\pi r^2$ with respect to $t$:

$$\frac{dS}{dt} = 8\pi r \cdot \frac{dr}{dt}$$

Substitute $r = 5$ and $\dfrac{dr}{dt} = 2$:

$$\frac{dS}{dt} = 8\pi(5)(2) = 80\pi \approx 251.3 \text{ cm}^2/\text{s}$$

Geometric insight. Notice that $\dfrac{dV}{dt} = 4\pi r^2 \cdot \dfrac{dr}{dt} = S \cdot \dfrac{dr}{dt}$. The surface area $S$ is literally the derivative $\dfrac{dV}{dr}$. Geometrically: a thin shell of thickness $dr$ on the outside of the sphere has volume $\approx S \cdot dr$, so $dV = S\,dr$. This relationship is exact, not an approximation.

The graph below shows $V(r)$ (blue) and $S(r)$ (green) as functions of $r$. Notice that the slope of the $V$-curve at any point equals the height of the $S$-curve at the same $r$ — confirming $\dfrac{dV}{dr} = S(r)$. Drag the slider to explore this at different radii.

Figure 4.1 — Sphere: V(r) and S(r) with dV/dr = S

$V(r) = \tfrac{4}{3}\pi r^3$ (blue) and $S(r) = 4\pi r^2$ (green). The dashed tangent to $V$ at $r = r_0$ has slope $= S(r_0)$ (the green dot's height). At $r = 5$: slope $= 100\pi \approx 314$, confirming $\frac{dV}{dr}\big|_{r=5} = S(5)$.

Example 5: Rotating Spotlight — $\tan\theta$ Related Rate

A spotlight is mounted at ground level, $6$ m from a straight wall. A person walks along the wall at $1.5$ m/s. At what rate is the spotlight rotating when the person is $8$ m from the point on the wall nearest to the spotlight?

Setup. Let:

• $x$ = distance of the person from the nearest point $N$ on the wall, measured along the wall (changes with time)
• $\theta$ = angle the spotlight beam makes with the perpendicular $SN$ from the spotlight $S$ to the wall
• $d = 6$ m = fixed distance from spotlight to wall

From the right triangle $SNP$:

$$\tan\theta = \frac{x}{6}$$

Differentiate with respect to $t$. Using the chain rule on both sides:

$$\sec^2\!\theta \cdot \frac{d\theta}{dt} = \frac{1}{6}\,\frac{dx}{dt}$$

Substitute known values at $x = 8$.

From $\tan\theta = \dfrac{8}{6} = \dfrac{4}{3}$, use $\sec^2\theta = 1 + \tan^2\!\theta$:

$$\sec^2\!\theta = 1 + \left(\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}$$

Now substitute $\sec^2\theta = \dfrac{25}{9}$ and $\dfrac{dx}{dt} = 1.5$:

$$\frac{25}{9}\cdot\frac{d\theta}{dt} = \frac{1}{6}(1.5) = \frac{1}{4}$$ $$\frac{d\theta}{dt} = \frac{1}{4}\cdot\frac{9}{25} = \frac{9}{100} = 0.09 \text{ rad/s}$$

Physical interpretation. As $x$ increases (person moves farther from $N$), $\sec^2\theta$ grows — so for the same walking speed $\dfrac{dx}{dt}$, the spotlight rotation $\dfrac{d\theta}{dt}$ slows down. The spotlight turns fastest when the person is directly in front (at $N$, where $\theta = 0$, $\sec^2\theta = 1$, giving $\dfrac{d\theta}{dt} = \dfrac{1.5}{6} = 0.25$ rad/s) and most slowly when the person is far down the wall.

The interactive diagram below shows the spotlight geometry. Drag the slider to move the person along the wall and watch how the beam angle $\theta$ changes — and how rapidly it changes per unit of the person's movement.

Figure 4.2 — Rotating Spotlight Geometry

Spotlight at $S = (0,0)$, wall at $x = 6$. The person $P$ is at $(6,\,x)$ — drag the slider to walk them along the wall. The amber beam rotates; the dashed blue segment shows the perpendicular $SN$; the purple arc marks angle $\theta$. At the default position $x=8$, $\theta \approx 53.1°$ and $d\theta/dt = 0.09$ rad/s.

Example 4: Approximating a Square Root

Use linear approximation to estimate $\sqrt{50}$.

Solution. Let $f(x) = \sqrt{x}$, with $a = 49$ (the nearest perfect square). Then:

$$f(49) = 7, \qquad f'(x) = \frac{1}{2\sqrt{x}} \implies f'(49) = \frac{1}{14}$$

With $\Delta x = 50 - 49 = 1$, the linearization gives:

$$\sqrt{50} \approx L(50) = 7 + \frac{1}{14}(1) \approx 7.0714$$

The actual value is $\sqrt{50} = 7.07107\ldots$, so the approximation is accurate to four significant figures.

Figure 4.4 — Square Root Approximation at x = 49

The blue curve is $y = \sqrt{x}$. The red dashed line is the tangent at $(49, 7)$. The green dot marks $\sqrt{50}$ on the curve; the orange dot marks the linear approximation $L(50) \approx 7.0714$. They are nearly coincident.

Example 5: Approximating a Cube Root — $y = \sqrt[3]{x}$

Use linear approximation to estimate $\sqrt[3]{8.1}$.

Solution. Let $f(x) = x^{1/3}$, with $a = 8$ (the nearest perfect cube). Compute:

$$f(8) = 2, \qquad f'(x) = \frac{1}{3}x^{-2/3} \implies f'(8) = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$$

The linearization at $a = 8$ is $L(x) = 2 + \frac{1}{12}(x - 8)$. With $\Delta x = 0.1$:

$$\sqrt[3]{8.1} \approx L(8.1) = 2 + \frac{0.1}{12} = 2 + 0.008\overline{3} \approx 2.0083$$

The actual value is $\sqrt[3]{8.1} = 2.00832\ldots$, matching to four decimal places. The error is only $0.00002$ — remarkably accurate because $f''(x) = -\frac{2}{9}x^{-5/3}$ is small near $x = 8$, so the curve is nearly straight there.

Figure 4.5 — Cube Root Approximation at x = 8

The blue curve is $y = x^{1/3}$. The red dashed line is $L(x) = 2 + \frac{1}{12}(x-8)$. The green dot marks $\sqrt[3]{8.1}$ on the curve; the orange dot marks $L(8.1) \approx 2.0083$. Drag the slider to zoom in: the two points become indistinguishable.

Example 6: Approximating a Fifth Power — $y = x^5$

Use linear approximation to estimate $(1.02)^5$.

Solution. Let $f(x) = x^5$, with $a = 1$. Then:

$$f(1) = 1, \qquad f'(x) = 5x^4 \implies f'(1) = 5$$

The linearization at $a = 1$ is $L(x) = 1 + 5(x - 1)$. With $\Delta x = 0.02$:

$$(1.02)^5 \approx L(1.02) = 1 + 5(0.02) = 1 + 0.10 = 1.10$$

The actual value is $(1.02)^5 = 1.10408\ldots$, so the error is about $0.004$. The approximation is accurate to two decimal places, but less accurate than the cube-root example because $y = x^5$ has larger curvature ($f''(1) = 20$) near $x = 1$, causing the tangent line to diverge faster.

Figure 4.6 — Fifth Power Approximation at x = 1

The blue curve is $y = x^5$. The red dashed line is the tangent at $(1,\,1)$: $L(x) = 1 + 5(x-1)$. The orange dot is the approximation $L(1.02) = 1.10$; the green dot is the true value $(1.02)^5 \approx 1.104$. The visible gap reflects the large curvature of $x^5$.

Example 7: Using Differentials to Estimate Error

The radius of a sphere is measured as $6$ cm with a possible error of $\pm 0.02$ cm. Estimate the maximum error in the calculated volume.

Solution. The volume is $V = \frac{4}{3}\pi r^3$, so the differential is:

$$dV = 4\pi r^2 \, dr$$

With $r = 6$ and $dr = \pm 0.02$:

$$dV = 4\pi(36)(0.02) = 2.88\pi \approx 9.05 \text{ cm}^3$$

The actual volume is $V = \frac{4}{3}\pi(216) = 288\pi \approx 904.8$ cm$^3$. The relative error is:

$$\frac{dV}{V} = \frac{2.88\pi}{288\pi} = 0.01 = 1\%$$

This equals $3 \cdot \frac{dr}{r} = 3 \cdot \frac{0.02}{6} = 1\%$. In general, for $V = \frac{4}{3}\pi r^3$, the relative error in volume is always three times the relative error in radius — a consequence of the exponent 3.

Figure 4.7 — Sphere Volume Differential: dV vs ΔV

The blue curve is $V(r) = \frac{4}{3}\pi r^3$. The red dashed line is the tangent at $r = 6$. Drag the $dr$ slider to change the measurement error: the orange segment shows $dV$ (tangent-line approximation) and the green segment shows the true $\Delta V$.

Example 8: Closed Interval Method

Find the absolute maximum and minimum of $f(x) = x^3 - 3x^2 + 1$ on $[-1, 4]$.

Solution. First, find critical points: $f'(x) = 3x^2 - 6x = 3x(x - 2)$. Setting $f'(x) = 0$ gives $x = 0$ and $x = 2$, both in $(-1, 4)$.

Evaluate $f$ at critical points and endpoints:

• $f(-1) = -1 - 3 + 1 = -3$
• $f(0) = 0 - 0 + 1 = 1$
• $f(2) = 8 - 12 + 1 = -3$
• $f(4) = 64 - 48 + 1 = 17$

The absolute maximum is $17$ at $x = 4$. The absolute minimum is $-3$, attained at both $x = -1$ and $x = 2$.

Example 9: First Derivative Test

Find and classify all local extrema of $g(x) = x^4 - 4x^3$.

Solution. Compute $g'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$. The critical points are $x = 0$ and $x = 3$.

Sign analysis of $g'$:

• For $x < 0$: $g'(-1) = 4(1)(-4) = -16 < 0$ (decreasing)
• For $0 < x < 3$: $g'(1) = 4(1)(-2) = -8 < 0$ (decreasing)
• For $x > 3$: $g'(4) = 4(16)(1) = 64 > 0$ (increasing)

At $x = 0$, $g'$ does not change sign (negative on both sides), so there is no local extremum. At $x = 3$, $g'$ changes from negative to positive, so $g$ has a local minimum of $g(3) = 81 - 108 = -27$.

Portrait 1 — The Function $f(x) = x^3 - x$

First, find the key points algebraically:

• Critical points: $f'(x) = 3x^2 - 1 = 0 \;\Rightarrow\; x = \pm\dfrac{1}{\sqrt{3}} \approx \pm 0.577$
• Inflection point: $f''(x) = 6x = 0 \;\Rightarrow\; x = 0$, giving the point $(0,\,0)$
• Local max at $x = -\tfrac{1}{\sqrt{3}}$: $f''\!\left(-\tfrac{1}{\sqrt{3}}\right) = -\tfrac{6}{\sqrt{3}} < 0$  ✓
• Local min at $x = \tfrac{1}{\sqrt{3}}$: $f''\!\left(\tfrac{1}{\sqrt{3}}\right) = \tfrac{6}{\sqrt{3}} > 0$  ✓

In the graph below, the curve is colored by concavity: orange = concave down ($x < 0$), green = concave up ($x > 0$). Drag the slider to move a tangent line along the curve — watch how the slope decreases through the orange region and increases through the green region.

Figure 4.8 — f(x) = x³ − x Colored by Concavity

$f(x) = x^3 - x$. Orange: concave down. Green: concave up. Purple dot: inflection point $(0, 0)$. Drag the slider $t$ to see how the tangent line's slope changes.

Portrait 2 — The First Derivative $f'(x) = 3x^2 - 1$

The value of $f'(x)$ at any point is the slope of $f$ there. Watch what $f'$ does as $x$ increases:

• $x < 0$ (orange): $f'$ is decreasing (the parabola descends toward its vertex). The slope of $f$ is shrinking — this means $f$ is concave down.
• $x = 0$: $f'$ reaches its minimum value of $-1$. The slope of $f$ stops falling and starts rising. This transition is exactly the inflection point of $f$.
• $x > 0$ (green): $f'$ is increasing (the parabola rises). The slope of $f$ is growing — this means $f$ is concave up.

The two zeros $x = \pm\tfrac{1}{\sqrt{3}}$ are where the slope of $f$ is zero — the local max and min of $f$.

Figure 4.9 — f′(x) = 3x² − 1 Colored by Direction

$f'(x) = 3x^2 - 1$. Orange (decreasing, $x < 0$): $f$ is concave down. Green (increasing, $x > 0$): $f$ is concave up. The minimum of $f'$ at $x = 0$ corresponds exactly to the inflection point of $f$.

Portrait 3 — The Second Derivative $f''(x) = 6x$

$f''(x) = 6x$ is a straight line through the origin. Its sign is all you need to know:

• $x < 0$: $f''(x) = 6x < 0$  →  concave down
• $x = 0$: $f''(0) = 0$ and the sign changes  →  inflection point at $(0,0)$
• $x > 0$: $f''(x) = 6x > 0$  →  concave up

The shaded regions below emphasize this: orange shading where $f'' < 0$ (concave down), green shading where $f'' > 0$ (concave up). The moment $f''$ crosses zero from negative to positive is exactly when $f$ transitions from dome to bowl.

Figure 4.10 — f′′(x) = 6x Sign and Concavity Regions

$f''(x) = 6x$. Below the $x$-axis (orange): $f$ is concave down. Above the $x$-axis (green): $f$ is concave up. The zero-crossing at $x = 0$ is the inflection point of $f$.

Example 10: Concavity and Inflection Points

Find the intervals of concavity and all inflection points of $g(x) = x^3 - 6x^2 + 9x + 2$.

Solution. Compute derivatives:

$$g'(x) = 3x^2 - 12x + 9, \qquad g''(x) = 6x - 12 = 6(x - 2)$$

Setting $g''(x) = 0$ gives $x = 2$. Sign analysis:

• $x < 2$: $g''(1) = 6(1-2) = -6 < 0$  →  concave down
• $x > 2$: $g''(3) = 6(3-2) = 6 > 0$  →  concave up

The sign changes at $x = 2$, so there is an inflection point at $(2,\;g(2)) = (2,\;4)$. Notice the structure is identical to our walkthrough function: $g''(x) = 6(x-2)$ is just $f''(x) = 6x$ shifted right by $2$ units, so the inflection point shifts from $(0,0)$ to $(2,4)$.

Example 11: Second Derivative Test

Find and classify the critical points of $h(x) = 3x^5 - 5x^3$ using the Second Derivative Test.

Solution.

$$h'(x) = 15x^4 - 15x^2 = 15x^2(x^2 - 1) = 15x^2(x-1)(x+1)$$

Critical points: $x = -1,\; 0,\; 1$.

$$h''(x) = 60x^3 - 30x = 30x(2x^2 - 1)$$

Apply the Second Derivative Test:

• $h''(-1) = 30(-1)(2-1) = -30 < 0$  →  local maximum at $(-1,\;2)$
• $h''(0) = 0$  →  inconclusive. By the First Derivative Test, $h'$ is negative on both sides of $x=0$, so there is no extremum at $x = 0$.
• $h''(1) = 30(1)(2-1) = 30 > 0$  →  local minimum at $(1,\;-2)$

Example 12: Maximize Enclosed Area

A farmer has $600$ meters of fencing and wants to enclose a rectangular field that borders a straight river. No fencing is needed along the river. What dimensions maximize the enclosed area?

Solution. Let $x$ be the length of the side perpendicular to the river and $y$ the length of the side parallel to the river. The fencing constraint is:

$$2x + y = 600 \quad \Longrightarrow \quad y = 600 - 2x$$

The area to maximize is:

$$A(x) = x \cdot y = x(600 - 2x) = 600x - 2x^2$$

The domain is $0 \leq x \leq 300$ (since $y \geq 0$). Differentiate:

$$A'(x) = 600 - 4x$$

Setting $A'(x) = 0$: $x = 150$. Since $A''(x) = -4 < 0$, this is a maximum. Then $y = 600 - 300 = 300$.

The maximum area is $A(150) = 150 \times 300 = 45{,}000$ m$^2$, with the field being $150$ m by $300$ m.

Example 13: Minimize Cost

An open-top rectangular box with a square base must have a volume of $32{,}000$ cm$^3$. The material for the base costs $\$10$ per cm$^2$ and the material for the sides costs $\$5$ per cm$^2$. Find the dimensions that minimize the total cost.

Solution. Let the base have side length $x$ and the height be $h$. The volume constraint is:

$$x^2 h = 32{,}000 \quad \Longrightarrow \quad h = \frac{32{,}000}{x^2}$$

The cost function is (base area times $\$10$) + (four sides times $\$5$):

$$C(x) = 10x^2 + 5 \cdot 4xh = 10x^2 + 20x \cdot \frac{32{,}000}{x^2} = 10x^2 + \frac{640{,}000}{x}$$

The domain is $x > 0$. Differentiate:

$$C'(x) = 20x - \frac{640{,}000}{x^2}$$

Set $C'(x) = 0$:

$$20x = \frac{640{,}000}{x^2} \quad \Longrightarrow \quad 20x^3 = 640{,}000 \quad \Longrightarrow \quad x^3 = 32{,}000 \quad \Longrightarrow \quad x = \sqrt[3]{32{,}000}$$

Since $32{,}000 = 32 \times 1000$, we get $x = \sqrt[3]{32} \times 10 = 2\sqrt[3]{4} \times 10 \approx 31.75$ cm.

Verify: $C''(x) = 20 + \frac{1{,}280{,}000}{x^3} > 0$ for all $x > 0$, confirming a minimum. The height is $h = \frac{32{,}000}{x^2} \approx 31.75$ cm. The box is (approximately) a cube with side $\approx 31.75$ cm.

Example 14: AP-Style Optimization

A particle moves along the $x$-axis so that its velocity at time $t \geq 0$ is given by $v(t) = t^2 - 6t + 8$. For what value of $t$ does the particle achieve its minimum velocity, and what is the particle's position at that time if $x(0) = 3$?

Solution. The velocity is $v(t) = t^2 - 6t + 8$. To minimize velocity, find where $v'(t) = 0$:

$$v'(t) = 2t - 6 = 0 \quad \Longrightarrow \quad t = 3$$

Since $v''(t) = 2 > 0$, this is a minimum. The minimum velocity is $v(3) = 9 - 18 + 8 = -1$ (the particle is moving to the left at speed 1).

For the position, integrate the velocity:

$$x(t) = \int_0^t v(s)\,ds + x(0) = \int_0^t (s^2 - 6s + 8)\,ds + 3$$ $$x(t) = \left[\frac{s^3}{3} - 3s^2 + 8s\right]_0^t + 3 = \frac{t^3}{3} - 3t^2 + 8t + 3$$

At $t = 3$:

$$x(3) = \frac{27}{3} - 27 + 24 + 3 = 9 - 27 + 24 + 3 = 9$$

Example 15: Jump Discontinuity — Maximum May Not Be Achieved

Find the absolute maximum and minimum of the piecewise function on $[0, 4]$:

$$f(x) = \begin{cases} -x^2 + 4x & 0 \leq x < 2 \\ x - 3 & 2 \leq x \leq 4 \end{cases}$$

Naïve approach (incomplete). Set $f'=0$ on each piece: Piece 1 gives $x=2$ (not in $[0,2)$, excluded); Piece 2 gives $f'=1 \neq 0$. Check endpoints: $f(0)=0$, $f(4)=1$. Conclusion: "max = 1, min = 0." This misses critical information.

Correct analysis. The function has a jump discontinuity at $x=2$. Always check what happens there:

• Actual value: $f(2) = 2-3 = -1$
• Left-hand limit: $\displaystyle\lim_{x \to 2^-} f(x) = -4+8 = 4$ — the curve approaches 4 but never reaches it (open circle)

All candidates: $f(0)=0$, $f(4)=1$, $f(2)=-1$, and the unachieved limit 4.

• Absolute minimum $= f(2) = -1$ ✓ (achieved)
• No absolute maximum: the supremum is 4, but $f(x) < 4$ for every $x \in [0,4]$ — the value 4 is never attained

The EVT requires continuity. Without it, a max may simply not exist even on a closed interval.

Figure 4.13 — Piecewise Function with Jump Discontinuity

Piecewise function with a jump at $x=2$. The orange open circle at $(2,\,4)$ marks the unachieved limit; the red filled dot at $(2,\,-1)$ is the actual value $f(2)$. The dashed line $y=4$ marks the supremum that is never attained — so no absolute maximum exists.

Example 16: No Interior Critical Points — Endpoints Are the Extrema

Find the absolute maximum and minimum of $\displaystyle f(x) = \frac{1}{1+x^2}$ on $[1,\,3]$.

Solution. $f$ is continuous on $[1,3]$, so the EVT guarantees both a max and a min exist. Find critical points:

$$f'(x) = \frac{-2x}{(1+x^2)^2}$$

Since $x > 0$ on $(1,3)$, we have $f'(x) < 0$ everywhere in the interior — the function is strictly decreasing. No critical points in $(1,3)$.

When there are no interior critical points, the extrema must be at the endpoints:

• $f(1) = \dfrac{1}{1+1} = \dfrac{1}{2}$  →  absolute maximum (left endpoint) ✓
• $f(3) = \dfrac{1}{1+9} = \dfrac{1}{10}$  →  absolute minimum (right endpoint) ✓

Yes, closed-interval endpoints fully qualify as locations of absolute extrema. The Closed Interval Method always evaluates endpoints for precisely this reason: a monotone function on $[a,b]$ will have its extrema at the endpoints, not anywhere inside.

Figure 4.14 — Endpoint Extrema: f(x) = 1/(1+x²) on [1, 3]

$f(x) = 1/(1+x^2)$ (gray dashed: full curve; blue: restricted to $[1,3]$). The function is strictly decreasing with no interior critical points. The green dot at $(1,\,\tfrac{1}{2})$ is the absolute maximum; the red dot at $(3,\,\tfrac{1}{10})$ is the absolute minimum — both at endpoints.

Example 17: Vertical Asymptote Inside the Interval

Does $\displaystyle f(x) = \frac{1}{x-1}$ have an absolute maximum or minimum on $[0,\,3]$?

Analysis. The function has a vertical asymptote at $x=1$, which lies inside $[0,3]$. Since $f$ is undefined at $x=1$, it cannot be continuous on $[0,3]$, and the EVT does not apply.

Examine the behavior near the asymptote:

• $x \to 1^-$: $f(x) \to -\infty$  →  no lower bound on $[0,1)$, so no absolute minimum
• $x \to 1^+$: $f(x) \to +\infty$  →  no upper bound on $(1,3]$, so no absolute maximum

Neither an absolute max nor an absolute min exists on $[0,3]$.

How to handle it in practice. Restrict the domain to a sub-interval that avoids the asymptote:

• On $[0,\;0.5]$: $f$ is continuous and decreasing ($f' = -1/(x-1)^2 < 0$). Absolute max $= f(0) = -1$; absolute min $= f(0.5) = -2$.
• On $[2,\;3]$: $f$ is continuous and decreasing. Absolute max $= f(2) = 1$; absolute min $= f(3) = \tfrac{1}{2}$.

Rule of thumb: Before applying optimization, always check that the objective function has no vertical asymptote (or any other discontinuity) inside the domain you are given.

Figure 4.15 — Vertical Asymptote Inside [0, 3]

$f(x) = 1/(x-1)$ on $[0,3]$. The red dashed line at $x=1$ is the vertical asymptote. The function is unbounded below on $[0,1)$ and unbounded above on $(1,3]$ — no absolute extrema exist on the full interval $[0,3]$.

Example 18: Applying the Mean Value Theorem

Let $f(x) = x^3 - x$ on the interval $[0, 2]$. Verify that the hypotheses of the MVT are satisfied and find all values of $c$ guaranteed by the theorem.

Solution. The function $f(x) = x^3 - x$ is a polynomial, so it is continuous on $[0, 2]$ and differentiable on $(0, 2)$. The hypotheses are satisfied.

Compute the average rate of change:

$$\frac{f(2) - f(0)}{2 - 0} = \frac{(8 - 2) - (0)}{2} = \frac{6}{2} = 3$$

We need $f'(c) = 3$. Since $f'(x) = 3x^2 - 1$:

$$3c^2 - 1 = 3 \quad \Longrightarrow \quad 3c^2 = 4 \quad \Longrightarrow \quad c^2 = \frac{4}{3} \quad \Longrightarrow \quad c = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 1.155$$

We discard the negative root since $c$ must lie in $(0, 2)$. The value $c = \frac{2\sqrt{3}}{3}$ is the point where the tangent line is parallel to the secant line.