MechForge · Calculus · Chapter 4 — Applications of Differentiation

4.1

Maximum and Minimum Values

Many practical problems reduce to finding the best possible outcome: the cheapest can, the greatest area, the fastest route. Each of these is an optimization problem, and each comes down to locating the maximum or minimum value of a function. We begin by defining precisely what those extreme values are.

Absolute (Global) Extrema

A function $f$ has an absolute maximum at $c$ if $f(c)\ge f(x)$ for every $x$ in the domain; the number $f(c)$ is the maximum value. It has an absolute minimum at $c$ if $f(c)\le f(x)$ for every $x$ in the domain. Together these are the extreme values of $f$.

Local (Relative) Extrema

A function $f$ has a local maximum at $c$ if $f(c)\ge f(x)$ for all $x$ in some open interval containing $c$, and a local minimum at $c$ if $f(c)\le f(x)$ nearby. A local extreme value is "best" only in its immediate neighborhood, not necessarily over the whole domain.

Figure 4.1.1 A function whose absolute minimum occurs at the endpoint $a$ and whose absolute maximum occurs at the interior point $d$. Interior peaks and valleys (at $b$ and $e$) are local extrema.

TheoremExtreme Value Theorem

If $f$ is continuous on a closed interval $[a,b]$, then $f$ attains an absolute maximum value $f(c)$ and an absolute minimum value $f(d)$ at some numbers $c$ and $d$ in $[a,b]$.

Both hypotheses matter. A discontinuous function on a closed interval, or a continuous function on an open interval, may fail to attain an extreme value, so neither continuity nor closedness can be dropped.

TheoremFermat's Theorem

If $f$ has a local maximum or minimum at $c$, and if $f'(c)$ exists, then $f'(c)=0$.

The converse is false: $f'(c)=0$ does not guarantee an extreme value (consider $f(x)=x^3$ at the origin), and an extreme value can occur where $f'$ fails to exist (consider $f(x)=|x|$ at the origin). Fermat's Theorem therefore tells us only where to look.

Critical Number

A critical number of $f$ is a number $c$ in the domain of $f$ at which either $f'(c)=0$ or $f'(c)$ does not exist. By Fermat's Theorem, every local extremum occurs at a critical number — though not every critical number produces an extremum.

Example 1 Finding critical numbers

Find the critical numbers of $f(x)=x^{3/5}(4-x)$.

Solution

Differentiate with the Product Rule and combine over a common denominator: \[ f'(x)=x^{3/5}(-1)+(4-x)\cdot\tfrac{3}{5}x^{-2/5}=\frac{12-8x}{5x^{2/5}}. \]

Set the numerator to zero: $12-8x=0\Rightarrow x=\tfrac{3}{2}$.

The derivative fails to exist where the denominator vanishes: $x=0$.

The critical numbers are $x=\tfrac{3}{2}$ and $x=0$.

To find the absolute extrema of a continuous function on a closed interval, we need only compare values at the critical numbers and at the endpoints.

ProcedureThe Closed Interval Method

To find the absolute maximum and minimum of a continuous $f$ on $[a,b]$:

Evaluate $f$ at every critical number in $(a,b)$.
Evaluate $f$ at the endpoints $a$ and $b$.
The largest of these values is the absolute maximum; the smallest is the absolute minimum.

Example 2 Closed Interval Method

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

Solution

$f'(x)=3x^2-6x=3x(x-2)$, so the critical numbers are $x=0$ and $x=2$, both inside the interval.

Values at critical numbers: $f(0)=1$ and $f(2)=-3$.

Values at endpoints: $f\!\left(-\tfrac12\right)=\tfrac18$ and $f(4)=17$.

Comparing the four numbers: the absolute maximum is $f(4)=17$ (at an endpoint) and the absolute minimum is $f(2)=-3$ (at a critical number).

Example 3 Extreme acceleration of a space shuttle

A model for the velocity of a launching shuttle (in ft/s) is $v(t)=0.001302\,t^3-0.09029\,t^2+23.61\,t-3.083$ for $0\le t\le 126$ s. Estimate the extreme values of the acceleration over this interval.

Solution

Acceleration is the derivative of velocity: \[ a(t)=v'(t)=0.003906\,t^2-0.18058\,t+23.61. \]

Apply the Closed Interval Method to $a$. Its derivative $a'(t)=0.007812\,t-0.18058$ vanishes at $t_1=\dfrac{0.18058}{0.007812}\approx 23.12$.

Evaluate: $a(0)=23.61$, $a(t_1)\approx 21.52$, $a(126)\approx 62.87$.

Maximum acceleration $\approx 62.87\ \text{ft/s}^2$; minimum acceleration $\approx 21.52\ \text{ft/s}^2$.

Instructor Note

In every real design problem, check the endpoints first — students lose marks by hunting only for $f'(x)=0$ and forgetting that the global optimum of a bounded design (a beam length, a tank radius constrained to a range) very often sits right on the boundary, exactly as the maximum did in Example 2.

4.2

The Mean Value Theorem

Much of this chapter rests on a single central result that connects a function's average rate of change to its instantaneous rate of change. To reach it we first need a special case due to Rolle.

TheoremRolle's Theorem

Let $f$ satisfy three hypotheses: (1) $f$ is continuous on $[a,b]$; (2) $f$ is differentiable on $(a,b)$; (3) $f(a)=f(b)$. Then there exists a number $c$ in $(a,b)$ with $f'(c)=0$.

Geometrically, if a smooth curve returns to its starting height, it must level off somewhere between — there is at least one horizontal tangent.

Example 1 Uniqueness of a root

Prove that $x^3+x-1=0$ has exactly one real root.

Solution

Existence (Intermediate Value Theorem): with $f(x)=x^3+x-1$, $f(0)=-1<0$ and $f(1)=1>0$, so a root lies between 0 and 1.

Uniqueness (Rolle, by contradiction): if there were two roots $a$ and $b$, then $f(a)=f(b)=0$ would force some $c$ with $f'(c)=0$.

But $f'(x)=3x^2+1\ge 1$ for all $x$, so $f'$ is never zero — a contradiction. Hence the root is unique.

TheoremThe Mean Value Theorem

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists a number $c$ in $(a,b)$ such that

Mean Value Theorem — two equivalent forms

\[ f'(c)=\frac{f(b)-f(a)}{b-a} \qquad\Longleftrightarrow\qquad f(b)-f(a)=f'(c)(b-a). \]

The slope of the secant line through $\big(a,f(a)\big)$ and $\big(b,f(b)\big)$ equals the slope of the tangent at some interior point — the tangent there is parallel to the secant.

Figure 4.2.1 The Mean Value Theorem guarantees an interior point $c$ where the tangent line is parallel to the secant joining the endpoints.

Example 2 Bounding a function from its derivative

Suppose $f(0)=-3$ and $f'(x)\le 5$ for all $x$. How large can $f(2)$ be?

Solution

Apply the MVT on $[0,2]$: there is a $c$ with $f(2)-f(0)=f'(c)(2-0)$, so $f(2)=-3+2f'(c)$.

Since $f'(c)\le 5$, we get $f(2)\le -3+2(5)=7$.

The largest possible value of $f(2)$ is $\boxed{7}$.

TheoremConstant Function Theorem

If $f'(x)=0$ for all $x$ in an interval $(a,b)$, then $f$ is constant on $(a,b)$.

Corollary. If $f'(x)=g'(x)$ for all $x$ in $(a,b)$, then $f-g$ is constant; that is, $f(x)=g(x)+c$. This corollary underlies the entire theory of antiderivatives in Section 4.9.

Instructor Note

The MVT is the engineer's bridge between data and rates. If a vehicle covers 180 km in 2 hours, the speedometer must have read 90 km/h at some instant. Whenever you only know endpoints but need to argue about an instantaneous rate, this is the theorem to reach for.

4.3

How Derivatives Affect the Shape of a Graph

The sign of $f'$ tells us whether a function rises or falls; the sign of $f''$ tells us how it bends. Together they reveal the complete shape of a graph.

What $f'$ says about $f$

TestIncreasing / Decreasing Test (I/D Test)

If $f'(x)>0$ on an interval, then $f$ is increasing there. If $f'(x)<0$ on an interval, then $f$ is decreasing there.

TestThe First Derivative Test

Suppose $c$ is a critical number of a continuous function $f$.

If $f'$ changes from positive to negative at $c$, then $f$ has a local maximum at $c$.
If $f'$ changes from negative to positive at $c$, then $f$ has a local minimum at $c$.
If $f'$ does not change sign at $c$, then $f$ has no local extremum there.

Example 1 Increase, decrease, and local extrema

For $f(x)=3x^4-4x^3-12x^2+5$, find where $f$ increases and decreases and identify its local extrema.

Solution

$f'(x)=12x^3-12x^2-24x=12x(x-2)(x+1)$, giving critical numbers $x=-1,\,0,\,2$.

A sign chart of the three factors shows: $f$ decreasing on $(-\infty,-1)$, increasing on $(-1,0)$, decreasing on $(0,2)$, increasing on $(2,\infty)$.

By the First Derivative Test: local minimum $f(-1)=0$, local maximum $f(0)=5$, local minimum $f(2)=-27$.

What $f''$ says about $f$

Concavity

A graph is concave upward on an interval if it lies above all of its tangent lines there, and concave downward if it lies below them. Equivalently, concave upward means the slope $f'$ is increasing; concave downward means $f'$ is decreasing.

Inflection Point

A point $P$ on a curve is an inflection point if the curve is continuous there and changes its direction of concavity at $P$ (from up to down or down to up). Candidate inflection points occur where $f''=0$ or where $f''$ fails to exist.

TestConcavity Test

If $f''(x)>0$ for all $x$ in an interval, the graph of $f$ is concave upward there. If $f''(x)<0$ throughout an interval, the graph is concave downward there.

TestThe Second Derivative Test

Suppose $f''$ is continuous near $c$.

If $f'(c)=0$ and $f''(c)>0$, then $f$ has a local minimum at $c$.
If $f'(c)=0$ and $f''(c)<0$, then $f$ has a local maximum at $c$.

The test is inconclusive when $f''(c)=0$ or $f''(c)$ does not exist; fall back on the First Derivative Test.

Example 2 Concavity, inflection, and extrema together

Discuss $y=x^4-4x^3$ with respect to concavity, points of inflection, and local extrema.

Solution

$f'(x)=4x^3-12x^2=4x^2(x-3)$, so $f'=0$ at $x=0$ and $x=3$.

$f''(x)=12x^2-24x=12x(x-2)$. At the critical numbers: $f''(3)=36>0\Rightarrow$ local minimum $f(3)=-27$; $f''(0)=0\Rightarrow$ test inconclusive.

Use the First Derivative Test at 0: $f'<0$ on both sides of 0, so there is no extremum there.

$f''=0$ at $x=0$ and $x=2$, and $f''$ changes sign at each, giving inflection points $(0,0)$ and $(2,-16)$.

Figure 4.3.1 The quartic $y=x^4-4x^3$: a local minimum at $(3,-27)$ and inflection points at $(0,0)$ and $(2,-16)$ where concavity reverses.

Instructor Note

When $f''(c)=0$ the Second Derivative Test is silent — it does not mean "no extremum." Switch to the First Derivative Test (the sign change of $f'$) rather than guessing. In practice the First Derivative Test is more reliable and often faster, even where both apply.

4.4

Indeterminate Forms and l'Hospital's Rule

A limit of a quotient where both parts approach 0, or both approach $\infty$, cannot be evaluated by the quotient law — the outcome is a genuine contest between numerator and denominator. l'Hospital's Rule resolves these indeterminate forms systematically.

Indeterminate Forms of Type $\tfrac{0}{0}$ and $\tfrac{\infty}{\infty}$

A limit $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}$ is an indeterminate form of type $\tfrac{0}{0}$ if both $f(x)\to 0$ and $g(x)\to 0$, and of type $\tfrac{\infty}{\infty}$ if both grow without bound. The value, if it exists, is not determined by the form alone.

Rulel'Hospital's Rule

Suppose $f$ and $g$ are differentiable with $g'(x)\ne 0$ near $a$ (except possibly at $a$), and the quotient is of type $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$. Then

l'Hospital's Rule

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

provided the limit on the right exists (or is $\pm\infty$). The rule also holds for one-sided limits and limits at $\pm\infty$. Differentiate numerator and denominator separately — this is not the Quotient Rule.

Example 1 A basic $\tfrac{0}{0}$ limit

Evaluate $\displaystyle\lim_{x\to 1}\frac{\ln x}{x-1}$.

Solution

Both $\ln x\to 0$ and $x-1\to 0$, so the form is $\tfrac{0}{0}$.

\[ \lim_{x\to 1}\frac{\ln x}{x-1}=\lim_{x\to 1}\frac{1/x}{1}=1. \]

Example 2 Repeated application

Calculate $\displaystyle\lim_{x\to\infty}\frac{e^x}{x^2}$.

Solution

Type $\tfrac{\infty}{\infty}$: $\displaystyle\lim_{x\to\infty}\frac{e^x}{x^2}=\lim_{x\to\infty}\frac{e^x}{2x}$.

Still $\tfrac{\infty}{\infty}$; apply again: $\displaystyle\lim_{x\to\infty}\frac{e^x}{2x}=\lim_{x\to\infty}\frac{e^x}{2}=\infty$.

This confirms that exponential growth outpaces any power of $x$.

Other Indeterminate Forms

Products of type $0\cdot\infty$, differences of type $\infty-\infty$, and powers of type $0^0$, $\infty^0$, and $1^\infty$ are first rewritten — as a quotient, over a common denominator, or via the identity $\big[f(x)\big]^{g(x)}=e^{\,g(x)\ln f(x)}$ — to produce a $\tfrac{0}{0}$ or $\tfrac{\infty}{\infty}$ form to which the rule applies.

Example 3 An indeterminate product, $0\cdot\infty$

Evaluate $\displaystyle\lim_{x\to 0^+} x\ln x$.

Solution

As $x\to 0^+$, $x\to 0$ while $\ln x\to-\infty$: type $0\cdot\infty$. Rewrite as a quotient: \[ x\ln x=\frac{\ln x}{1/x}. \]

Now type $\tfrac{-\infty}{\infty}$; apply the rule: \[ \lim_{x\to 0^+}\frac{\ln x}{1/x}=\lim_{x\to 0^+}\frac{1/x}{-1/x^2}=\lim_{x\to 0^+}(-x)=0. \]

Instructor Note

Always confirm the form before differentiating. A classic trap: $\displaystyle\lim_{x\to\pi^-}\frac{\sin x}{1-\cos x}$ looks like a candidate, but the denominator approaches $2$, not $0$ — the limit is simply $0/2=0$. Blindly applying the rule gives the wrong answer. Check, then differentiate.

4.5

Summary of Curve Sketching

Sketching a curve by hand assembles everything from earlier chapters and sections — domain, symmetry, asymptotes, increase and decrease, concavity — into one disciplined checklist. Calculus reveals features (exact extrema, inflection points) that a graphing device alone can easily hide.

ProcedureGuidelines for Sketching a Curve

Domain. Determine the set of $x$ for which $f$ is defined.
Intercepts. The $y$-intercept is $f(0)$; find $x$-intercepts by solving $f(x)=0$.
Symmetry. Test for even ($f(-x)=f(x)$), odd ($f(-x)=-f(x)$), or periodic behavior.
Asymptotes. Horizontal from $\lim_{x\to\pm\infty}f$; vertical where $f\to\pm\infty$; slant when the numerator degree exceeds the denominator by one.
Intervals of increase/decrease. Use the sign of $f'$ (I/D Test).
Local extrema. Apply the First (or Second) Derivative Test at critical numbers.
Concavity and inflection points. Use the sign of $f''$.
Sketch. Draw asymptotes, plot key points, and connect respecting all of the above.

Slant (Oblique) Asymptote

The line $y=mx+b$ is a slant asymptote of $y=f(x)$ if $\displaystyle\lim_{x\to\infty}\big[f(x)-(mx+b)\big]=0$. For a rational function whose numerator has degree exactly one more than the denominator, long division produces the equation of the slant asymptote directly.

Example 1 A rational curve with vertical and horizontal asymptotes

Sketch $y=\dfrac{2x^2}{x^2-1}$.

Solution

Domain: all $x$ with $x\ne\pm 1$. Intercepts: both are 0. Symmetry: $f(-x)=f(x)$, so the curve is even (symmetric about the $y$-axis).

Asymptotes: $\displaystyle\lim_{x\to\pm\infty}\frac{2x^2}{x^2-1}=2$, so $y=2$ is horizontal; the lines $x=1$ and $x=-1$ are vertical asymptotes.

Increase/decrease: $f'(x)=\dfrac{-4x}{(x^2-1)^2}$, so $f$ increases for $x<0$ and decreases for $x>0$ (excluding $\pm1$).

Extremum: the only critical number is $x=0$; $f'$ changes $+\to-$, giving a local maximum $f(0)=0$.

Concavity: $f''(x)=\dfrac{12x^2+4}{(x^2-1)^3}$; concave up for $|x|>1$, concave down for $|x|<1$; no inflection point ($\pm1$ are excluded from the domain).

Example 2 A curve with a slant asymptote

Sketch $y=\dfrac{x^3}{x^2+1}$.

Solution

Domain: all real $x$; symmetry: odd, symmetric about the origin; intercepts at the origin.

Slant asymptote: long division gives $f(x)=x-\dfrac{x}{x^2+1}\to x$, so $y=x$ is a slant asymptote.

Increase: $f'(x)=\dfrac{x^2(x^2+3)}{(x^2+1)^2}\ge 0$, so $f$ is increasing everywhere; though $f'(0)=0$, $f'$ does not change sign, so there is no extremum.

Inflection: $f''(x)=\dfrac{2x(3-x^2)}{(x^2+1)^3}$ changes sign at $x=0$ and $x=\pm\sqrt3$, giving inflection points at $(0,0)$ and $\left(\pm\sqrt3,\pm\tfrac{3\sqrt3}{4}\right)$.

Instructor Note

Run the checklist in order and resist the urge to skip steps you "already see." Symmetry, in particular, halves your work: spotting that a function is even or odd means you only analyze half the domain and reflect. Calculators miss the small, exact features — that is precisely why we sketch by hand first.

4.6

Graphing with Calculus and Calculators

Here the order is reversed: we begin with a machine-produced graph and then use calculus to refine it, choosing viewing rectangles intelligently and confirming the features the device suggests. Calculus and the calculator work best in tandem.

The Role of the Viewing Rectangle

A single viewing window rarely shows everything. A window wide enough to capture end behavior often hides fine detail near the origin, while a close-up may miss large-scale structure. The strategy is to combine several windows, guided by calculus, rather than trusting any one automatic plot.

Confirming Features with Derivatives

Graphs of $f'$ and $f''$ pin down the exact locations of maxima, minima, and inflection points that a raw plot only hints at. Where $f'$ changes sign, $f$ has an extremum; where $f''$ changes sign, the concavity reverses. These derivative graphs make small or hidden features unmistakable.

Example 1 Hidden detail in a polynomial

Use the graphs of $f'$ and $f''$ to find all maxima, minima, and intervals of concavity for $f(x)=2x^6+3x^5+3x^3-2x^2$.

Solution

A broad window shows only the dominant $2x^6$ end behavior; a window near the origin reveals subtler structure.

$f'(x)=12x^5+15x^4+9x^2-4x$. Its sign changes locate an absolute minimum near $x\approx-1.62$ and, surprisingly, a local maximum at $x=0$ and a local minimum near $x\approx 0.35$ — features hidden in the first plot.

$f''(x)=60x^4+60x^3+18x-4$. Its sign changes give inflection points near $x\approx-1.23$ and $x\approx 0.19$; $f$ is concave up outside and concave down between them.

No single graph showed all features — two windows together gave the full picture.

Families of Functions

When a formula contains a parameter, graphing several members reveals how the graph evolves and exposes transitional values — values of the parameter at which the basic shape changes (for example, where the number of asymptotes, critical numbers, or inflection points jumps). Studying $f(x)=\dfrac{1}{x^2+2x+c}$ shows a transition at $c=1$: two vertical asymptotes for $c<1$, one at $c=1$, and none for $c>1$.

Instructor Note

Treat the calculator as a fast first draft, never the final word. The bullet points it shows can be off by whole features. Differentiate, find where $f'$ and $f''$ vanish, and you turn an approximate picture into exact, defensible results — which is what you'll be graded on and what a design review will demand.

4.7

Optimization Problems

The extreme-value methods of this chapter shine in applied problems: minimizing cost, maximizing area or profit, finding the quickest route. The hard part is usually translation — turning a verbal description into a function of one variable to be optimized.

ProcedureSteps in Solving Optimization Problems

Understand the problem: identify the unknown, the given data, and the conditions.
Draw a diagram and label it.
Introduce notation: name the quantity $Q$ to be optimized and the other variables.
Express $Q$ in terms of those variables.
Reduce to one variable using the constraints, then state the domain.
Optimize with the methods of Sections 4.1 and 4.3 (Closed Interval Method when the domain is closed).

TestFirst Derivative Test for Absolute Extreme Values

Let $c$ be a critical number of a continuous $f$ on an interval. If $f'(x)>0$ for all $xc$, then $f(c)$ is the absolute maximum. If $f'(x)<0$ for $x0$ for $x>c$, then $f(c)$ is the absolute minimum. This handles open or unbounded domains, where the Closed Interval Method does not apply.

Example 1 Largest rectangular field along a river

A farmer has 2400 ft of fencing to enclose a rectangular field bordering a straight river, needing no fence on the river side. What dimensions maximize the area?

Solution

Let $x$ be the depth (the two sides perpendicular to the river) and $y$ the width. Area $A=xy$, with constraint $2x+y=2400$.

Solve the constraint: $y=2400-2x$, so $A(x)=x(2400-2x)=2400x-2x^2$, for $0\le x\le 1200$.

$A'(x)=2400-4x=0\Rightarrow x=600$. Compare $A(0)=0$, $A(600)=720{,}000$, $A(1200)=0$.

Maximum area $720{,}000\ \text{ft}^2$, with depth $600$ ft and width $1200$ ft.

Example 2 Cheapest cylindrical can

Design a cylindrical can to hold 1 L (1000 cm³) of oil using the least metal. Find the dimensions.

Solution

Surface area $A=2\pi r^2+2\pi r h$, with volume constraint $\pi r^2 h=1000$, so $h=\dfrac{1000}{\pi r^2}$.

Substitute: $A(r)=2\pi r^2+\dfrac{2000}{r}$, for $r>0$.

$A'(r)=4\pi r-\dfrac{2000}{r^2}=\dfrac{4(\pi r^3-500)}{r^2}=0\Rightarrow r=\sqrt[3]{500/\pi}$.

Since $A'<0$ before and $A'>0$ after this value, it is the absolute minimum. Then $h=\dfrac{1000}{\pi r^2}=2r$: the optimal height equals the diameter.

Example 3 Maximizing revenue with a rebate

A store sells 200 units a week at $350. Each $10 rebate raises weekly sales by 20 units. What rebate maximizes revenue?

Solution

With $x$ units sold, the demand (price) function is $p(x)=350-\tfrac{10}{20}(x-200)=450-\tfrac12 x$.

Revenue $R(x)=x\,p(x)=450x-\tfrac12 x^2$, so $R'(x)=450-x=0\Rightarrow x=450$ (an absolute maximum, since $R$ is a downward parabola).

Corresponding price $p(450)=225$; the rebate is $350-225=125$. The store should offer a $125 rebate.

Instructor Note

Step 5 — reducing to one variable using the constraint — is where most of the marks live. Write the constraint equation explicitly before optimizing, and always state the domain. The "can with height equal to diameter" result in Example 2 is the classic engineering benchmark; real cans differ because manufacturing and material-waste costs add terms the pure-geometry model ignores.

4.8

Newton's Method

Most equations cannot be solved by formula. Newton's method (the Newton–Raphson method) approximates a root by repeatedly following the tangent line down to the axis — the engine inside most numerical root-finders.

The Idea

Starting from an approximation $x_n$, replace the curve by its tangent line at $\big(x_n,f(x_n)\big)$. Because a line is easy to solve, its $x$-intercept $x_{n+1}$ is taken as the next, usually better, approximation. Iterating produces a sequence that, under favorable conditions, converges rapidly to a root.

Newton's Method — iteration formula

\[ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}, \qquad f'(x_n)\ne 0. \]

Figure 4.8.1 Each step follows the tangent at $\big(x_n,f(x_n)\big)$ to its $x$-intercept $x_{n+1}$, driving the approximation toward the true root $r$.

Example 1 A cubic root by iteration

Starting with $x_1=2$, find the third approximation $x_3$ to the root of $x^3-2x-5=0$.

Solution

With $f(x)=x^3-2x-5$ and $f'(x)=3x^2-2$, the iteration is $x_{n+1}=x_n-\dfrac{x_n^3-2x_n-5}{3x_n^2-2}$.

$x_2=2-\dfrac{2^3-2(2)-5}{3(2)^2-2}=2.1$.

$x_3=2.1-\dfrac{(2.1)^3-2(2.1)-5}{3(2.1)^2-2}\approx 2.0946$, accurate to four decimal places.

Example 2 Computing a sixth root

Use Newton's method to approximate $\sqrt[6]{2}$ to eight decimal places.

Solution

Finding $\sqrt[6]{2}$ means solving $x^6-2=0$; take $f(x)=x^6-2$, $f'(x)=6x^5$, so $x_{n+1}=x_n-\dfrac{x_n^6-2}{6x_n^5}$.

Starting at $x_1=1$, successive values stabilize: $x_5\approx x_6\approx 1.12246205$.

Thus $\sqrt[6]{2}\approx 1.12246205$. Stop when consecutive approximations agree to the required precision.

Instructor Note

The method is powerful but not foolproof. It can fail or stall when $f'(x_n)$ is near zero (a nearly flat tangent throws the next guess far away) or when a poor starting point sends iterates outside the domain. Always sketch or plot first to choose a sensible $x_1$, and verify convergence by watching successive values settle.

4.9

Antiderivatives

Differentiation has an inverse problem: given a rate, recover the quantity. A physicist with a velocity wants position; an engineer with a leak rate wants total volume lost. The function whose derivative is the given one is its antiderivative — the gateway to integration.

Antiderivative

A function $F$ is an antiderivative of $f$ on an interval $I$ if $F'(x)=f(x)$ for all $x$ in $I$. Antiderivatives are not unique: adding any constant gives another.

TheoremGeneral Antiderivative

If $F$ is an antiderivative of $f$ on an interval $I$, then the most general antiderivative on $I$ is $F(x)+C$, where $C$ is an arbitrary constant. This follows from the Constant Function Corollary of Section 4.2: any two antiderivatives of $f$ differ by a constant.

Key antidifferentiation formulas

\[ \int x^{n}\,dx=\frac{x^{n+1}}{n+1}+C\ (n\ne -1), \qquad \int \frac{1}{x}\,dx=\ln|x|+C, \] \[ \int e^{x}\,dx=e^{x}+C, \quad \int \cos x\,dx=\sin x+C, \quad \int \sin x\,dx=-\cos x+C, \] \[ \int \sec^2 x\,dx=\tan x+C, \quad \int \frac{1}{1+x^2}\,dx=\tan^{-1}x+C. \]

Differential Equation and Initial Condition

An equation involving a function's derivatives is a differential equation; its general solution carries arbitrary constants. An initial condition — a known value such as $f(0)$ — fixes those constants and selects the one particular solution that fits.

Example 1 Recovering a function from a second derivative

Find $f$ if $f''(x)=12x^2+6x-4$, with $f(0)=4$ and $f(1)=1$.

Solution

Antidifferentiate once: $f'(x)=4x^3+3x^2-4x+C$.

Antidifferentiate again: $f(x)=x^4+x^3-2x^2+Cx+D$.

Apply $f(0)=4$: $D=4$. Apply $f(1)=1$: $1+1-2+C+4=1\Rightarrow C=-3$.

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

Example 2 Rectilinear motion under gravity

A ball is thrown upward at 48 ft/s from a cliff 432 ft above the ground. Taking $g=32\ \text{ft/s}^2$, find its height $s(t)$, the time of maximum height, and when it lands.

Solution

Acceleration $a(t)=-32$. Antidifferentiate: $v(t)=-32t+C$; since $v(0)=48$, $v(t)=-32t+48$.

Maximum height when $v(t)=0$: $t=1.5$ s.

Antidifferentiate again: $s(t)=-16t^2+48t+D$; since $s(0)=432$, $s(t)=-16t^2+48t+432$.

Lands when $s(t)=0$: solving $t^2-3t-27=0$ gives $t=\dfrac{3(1+\sqrt{13})}{2}\approx 6.9$ s.

Instructor Note

Never drop the constant of integration — it is not optional. In motion problems each antidifferentiation introduces one constant, and you need exactly as many initial conditions as constants to pin down the answer (two integrations of acceleration require both $v(0)$ and $s(0)$). This is the conceptual seed of the definite integral in the next chapter.

Applications of Differentiation

Maximum and Minimum Values

Absolute (Global) Extrema

Local (Relative) Extrema

Critical Number

The Mean Value Theorem

How Derivatives Affect the Shape of a Graph

What \(f'\) says about \(f\)

What \(f''\) says about \(f\)

Concavity

Inflection Point

Indeterminate Forms and l'Hospital's Rule

Indeterminate Forms of Type \(\tfrac{0}{0}\) and \(\tfrac{\infty}{\infty}\)

Other Indeterminate Forms

Summary of Curve Sketching

Slant (Oblique) Asymptote

Graphing with Calculus and Calculators

The Role of the Viewing Rectangle

Confirming Features with Derivatives

Families of Functions

Optimization Problems

Newton's Method

The Idea

Antiderivatives

Antiderivative

Differential Equation and Initial Condition