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Lesson 70 — Consolidation Term 7: Maxima, L'Hôpital, Taylor, Newton

Integrated Differential Calculus workshop applied: optimization, curve sketching, L'Hôpital's Rule, Taylor series, concavity, marginal analysis, kinematics, and Newton-Raphson. All techniques derive from local linearization.

Used in: Year 2 High School · Equivalent Japanese Math II/III Ch. 6–7 · Equivalent German Leistungskurs Differentialrechnung

f(x) \approx \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k

The Taylor series is the backbone of Term 7: it unifies optimization ( $f'(a)=0$ ), L'Hôpital's Rule (ratio of linear approximations), curve sketching (sign of $f''$ ), Newton-Raphson (tangent iteration), and marginal analysis (derivative as instantaneous rate of change). Every tool this term is a special case of this formula.

Choose your door

Rigorous notation, full derivation, hypotheses

Unified Theory: Linearization and Its Applications

The Mother Concept: Taylor Approximation

"The Taylor polynomial of degree $n$ centered at $x=a$ is the unique polynomial of degree $n$ that agrees with $f$ in value and in all its first $n$ derivatives at $x=a$ ." — Active Calculus §8.4

Optimization: Critical Points and the Second Derivative Test

"Finding the maximum and minimum values of a function also has practical significance because we can use this method to solve optimization problems, such as increasing profit, minimizing cost, and maximizing area." — OpenStax Calculus Vol. 1 §4.3

Concavity and Inflection Points

L'Hôpital's Rule: Ratio of Linearizations

"L'Hôpital's Rule applies whenever both $f(x) \to 0$ and $g(x) \to 0$ as $x \to a$ , or whenever $f(x) \to \pm\infty$ and $g(x) \to \pm\infty$ ." — OpenStax Calculus Vol. 1 §4.8

Idea via Taylor of Order 1. For $a = 0$ : $f(x) \approx f'(0) x$ and $g(x) \approx g'(0) x$ , so $f(x)/g(x) \approx f'(0)/g'(0)$ . L'Hôpital formalizes exactly this idea of a ratio of linearizations.

Newton-Raphson: Tangent Iteration

Newton-Raphson: the tangent to the curve at point x₀ crosses the axis at x₁, converging to the real root.

Unified Function Analysis Pipeline

Solved Examples

Example— 1· Optimization of an Open-Top Box (Application)

Problem: From a $24 \times 24$ cm sheet of cardboard, squares of side length $x$ are cut from the four corners, and the flaps are folded up to form an open-top box. Determine the value of $x$ that maximizes the volume.

Strategy: Express the volume as a function of $x$ , find the critical point via $V'(x) = 0$ , confirm it's a maximum using the second derivative test, and verify that $x$ is within the physical domain $0 < x < 12$ .

Solution:

Volume: $V(x) = x(24 - 2x)^2$ , domain: $x \in (0, 12)$ .
Expand: $V(x) = x(576 - 96x + 4x^2) = 4x^3 - 96x^2 + 576x$ .
Differentiate: $V'(x) = 12x^2 - 192x + 576 = 12(x^2 - 16x + 48) = 12(x-4)(x-12)$ .
Critical points in $(0,12)$ : $x = 4$ (the $x = 12$ is at the boundary, yielding zero volume).
Test: $V''(x) = 24x - 192$ ; $V''(4) = 96 - 192 = -96 < 0 \Rightarrow$ local maximum.
Maximum Volume: $V(4) = 4(24 - 8)^2 = 4 \cdot 256 = 1024\ \text{cm}^3$ .

Verification: $V(0) = 0$ , $V(12) = 0$ , $V(4) = 1024 > 0$ . The maximum is interior to the domain. Check: $V(4) = 4 \cdot 16^2 = 4 \cdot 256 = 1024$ .

Source. OpenStax Calculus Vol. 1 §4.7, ex. 4.41 — CC-BY-NC-SA license.

Example— 2· Indeterminate Limit via L'Hôpital's Rule (Application)

Problem: Calculate $\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$ .

Strategy: The form is $0/0$ at $x = 0$ . Apply L'Hôpital's Rule once and check if the new form is still $0/0$ ; if so, apply it again. Alternatively, use the Maclaurin series of $e^x$ to derive the result more quickly.

Solution:

Check form: $e^0 - 1 - 0 = 0$ and $0^2 = 0$ — $0/0$ indeterminate form.
1st L'Hôpital application: $\lim_{x \to 0} \frac{e^x - 1}{2x}$ Still $0/0$ (numerator $\to 0$ , denominator $\to 0$ ).
2nd application: $\lim_{x \to 0} \frac{e^x}{2} = \frac{e^0}{2} = \frac{1}{2}.$
Verification via Taylor: $e^x = 1 + x + x^2/2 + O(x^3)$ , so $e^x - 1 - x = x^2/2 + O(x^3)$ . Then $\frac{e^x - 1 - x}{x^2} \to \frac{x^2/2}{x^2} = \frac{1}{2}$ . Consistent.

Verification: For $x = 0.1$ : numerator $= e^{0.1} - 1.1 \approx 1.10517 - 1.1 = 0.00517$ ; denominator $= 0.01$ ; ratio $\approx 0.517 \approx 1/2$ .

Source. Active Calculus §2.6, exercise on indeterminate forms — CC-BY-NC-SA license.

Example— 3· Maclaurin Expansion and Limit (Intermediate)

Problem: Use the Maclaurin series of $\sin x$ to calculate $\displaystyle\lim_{x \to 0} \frac{\sin x - x + x^3/6}{x^5}$ .

Strategy: Expand $\sin x$ in Maclaurin series up to order 5, substitute into the numerator, and simplify — Taylor series is more elegant than applying L'Hôpital's Rule five times.

Solution:

Maclaurin series of $\sin x$ : $\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \frac{x^7}{5040} + \cdots$
Numerator: $\sin x - x + \frac{x^3}{6} = \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\right) - x + \frac{x^3}{6} = \frac{x^5}{120} + O(x^7).$
Limit: $\lim_{x \to 0} \frac{x^5/120 + O(x^7)}{x^5} = \frac{1}{120}.$

Verification: For $x = 0.1$ : $\sin(0.1) \approx 0.099833417$ ; $\sin x - x + x^3/6 = 0.099833417 - 0.1 + 0.000166667 = 0.000000083$ ; $x^5 = 10^{-5}$ ; ratio $\approx 0.0083 \approx 1/120 = 0.00833$ .

Source. Active Calculus §8.4 — CC-BY-NC-SA license.

Example— 4· Complete Sketch of a Rational Function (Intermediate)

Problem: Sketch $f(x) = \dfrac{x^2}{x^2 - 1}$ , identifying domain, asymptotes, extrema, and concavity.

Strategy: Follow the 7-step function analysis pipeline. The function is rational with easily identifiable denominator zeros.

Solution:

Domain: $x^2 - 1 \neq 0 \Rightarrow x \neq \pm 1$ . Domain: $\mathbb{R} \setminus \{-1, 1\}$ .
Intercepts/Symmetry: $f(0) = 0$ (y-intercept). Zeros: $x^2 = 0 \Rightarrow x = 0$ . $f(-x) = f(x)$ : even function (symmetric about the y-axis).
Asymptotes: vertical at $x = \pm 1$ ; horizontal: $\lim_{x\to\pm\infty} f(x) = 1$ (asymptote $y = 1$ ).
$f'$ : $f'(x) = \dfrac{-2x}{(x^2-1)^2}$ . Critical point: $x = 0$ . Signs: $f' > 0$ for $x < 0$ ; $f' < 0$ for $x > 0$ — local maximum at $x = 0$ .
$f''$ : $f''(x) = \dfrac{2(3x^2+1)}{(x^2-1)^3}$ . For $x \in (-1,1)$ : denominator $< 0$ , numerator $> 0$ , thus $f'' < 0$ (concave down). For $|x| > 1$ : $f'' > 0$ (concave up). No inflection points in the domain (sign change of $f''$ occurs at points outside the domain $x = \pm 1$ ).
Extremum: $f(0) = 0$ — local (and absolute) maximum in $(-1,1)$ .

Verification: $f(0) = 0 < 1 = \lim_{x\to\infty} f(x)$ makes sense: in the interval $(-1,1)$ the function goes from $0$ down to $-\infty$ at the boundaries and back up; the maximum is $f(0)=0$ in this interval.

Source. APEX Calculus §3.5, ex. 42 — CC-BY-NC license.

Example— 5· Newton-Raphson for a Transcendental Equation (Real Modeling)

Problem: An investment of R$ 10,000 grows at a monthly rate $r$ to be determined. After 12 months, the amount is R$ 11,268. Solve $f(r) = (1+r)^{12} - 1.1268 = 0$ using Newton-Raphson method, with $x_0 = 0.01$ (1% per month).

Strategy: $f'(r) = 12(1+r)^{11}$ . Iterate $x_{n+1} = x_n - f(x_n)/f'(x_n)$ until stabilization.

Solution:

Iteration 1: $f(0.01) = (1.01)^{12} - 1.1268 = 1.12683 - 1.1268 = 0.00003$ . $f'(0.01) = 12 \cdot (1.01)^{11} = 12 \cdot 1.11567 = 13.388$ . $x_1 = 0.01 - 0.00003/13.388 \approx 0.009998$ .

After 2 iterations, $r^* \approx 1.00\%$ per month (annual nominal rate $\approx 12.68\%$ p.a.).

Verification: $(1.01)^{12} = 1.12683 \approx 1.1268$ . Consistent with the problem statement.

Source. APEX Calculus §6.1, Newton-Raphson application — CC-BY-NC license.

Exercise list

40 exercises · 10 with worked solution (25%)

15 15 2 5 3

Ex. 70.1Answer key
Find the critical points, local extrema, and inflection point of $f(x) = x^3 - 6x^2 + 9x + 1$ .
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$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$ . Critical points: $x=1, x=3$ . $f''(x)=6x-12$ : $f''(1)=-6<0$ (local max), $f''(3)=6>0$ (local min). Inflection at $x=2$ ( $f''=0$ changes sign).
Show step-by-step (with the why)
1. Calculate $f'$ and factor. Why: critical points are where $f'=0$ .
2. Create a sign table for $f'$ : positive on $(-\infty,1)$ , negative on $(1,3)$ , positive on $(3,+\infty)$ .
3. Calculate $f''$ and evaluate at critical points. Why: second derivative test confirms the nature of the extremum.
4. Find inflection point: $f''(2)=0$ and $f''$ changes sign at $x=2$ .
Mnemonic: a cubic function with a positive leading coefficient always has a local maximum to the left and a local minimum to the right.
Ex. 70.2
Maximize $f(x) = xe^{-x}$ on $[0, +\infty)$ . What is the absolute maximum?
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$f'(x)=e^{-x} - xe^{-x} = e^{-x}(1-x)$ . Critical point: $x=1$ . $f''(1) = e^{-1}(1-1) + e^{-1}(-1) = -e^{-1} < 0$ : maximum. $f(1)=1/e$ . At $x=0$ : $f(0)=0$ . Absolute maximum at $x=1$ .
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1. Differentiate using the product rule: $f'(x) = 1\cdot e^{-x} + x\cdot(-e^{-x}) = e^{-x}(1-x)$ .
2. Since $e^{-x}>0$ always, the sign of $f'$ is the sign of $(1-x)$ . Critical point: $x=1$ .
3. Sign table: $f'>0$ on $(0,1)$ , $f'<0$ on $(1,\infty)$ — maximum at $x=1$ .
4. Compare with the boundary: $f(0)=0 < f(1)=1/e \approx 0.368$ and $f(x)\to 0$ as $x\to\infty$ . Thus $x=1$ is the absolute maximum.
Curiosity: the function $xe^{-x}$ models the Gamma probability distribution with parameter 2 — the mode of the distribution is exactly $x=1$ .
Ex. 70.3
Sketch $f(x) = (x^2 - 4)/x$ . Identify asymptotes, monotonicity, and concavity.
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$f(x)=(x^2-4)/x = x - 4/x$ . Vertical asymptote: $x=0$ . Slant asymptote: $y=x$ . $f'(x)=1+4/x^2>0$ always (no extrema). $f''(x)=-8/x^3$ : concave up on $x<0$ , concave down on $x>0$ . No zeros of the function in the domain.
Ex. 70.4
Find the absolute minimum of $f(x) = x + 4/x$ on $(0, +\infty)$ .
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$f'(x)=1-4/x^2=0 \Rightarrow x=2$ (in the domain $(0,\infty)$ ). $f''(x)=8/x^3>0$ : minimum. $f(2)=2+4/2=4$ .
Ex. 70.5Answer key
A cylindrical can with $V = 1000\ \text{cm}^3$ must be constructed using minimal material. Determine the optimal radius and height.
Solve online OpenStax Calculus Vol. 1 · §4.7
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Cylindrical can: radius $r$ , height $h$ , $V = \pi r^2 h = 1000$ cm³. Material $S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2000/r$ . $S'(r)=4\pi r - 2000/r^2=0 \Rightarrow r^3=500/\pi \Rightarrow r\approx 5.42$ cm; $h=2r\approx 10.84$ cm (height = diameter is the classic result).
Show step-by-step (with the why)
1. Variables: $r$ (radius) and $h$ (height). Objective: minimize total area $S$ .
2. Constraint: $\pi r^2 h = 1000$ , so $h = 1000/(\pi r^2)$ .
3. Substitute into $S$ : $S(r) = 2\pi r^2 + 2\pi r \cdot 1000/(\pi r^2) = 2\pi r^2 + 2000/r$ .
4. Differentiate and set to zero: $S'(r)=4\pi r - 2000/r^2=0 \Rightarrow 4\pi r^3=2000$ .
5. Confirm minimum: $S''(r)=4\pi + 4000/r^3 > 0$ always.
Observation: universal result — a can of fixed volume with minimal material always has $h = 2r$ (height equals diameter). Real beverage cans deviate due to manufacturing constraints.
Ex. 70.6
From a $20 \times 30\ \text{cm}$ piece of cardboard, squares of side $x$ are cut from the corners, and the flaps are folded up. What value of $x$ maximizes the volume of the box?
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$V(x) = x(20-2x)(30-2x)$ , $x \in (0,10)$ . $V'(x) = 12x^2 - 200x + 600 = 4(3x^2-50x+150)=4(3x-20)(x-...)$ . Roots: $x=(50\pm\sqrt{2500-1800})/6 = (50\pm \sqrt{700})/6$ . $\sqrt{700}\approx 26.46$ . Roots: $x\approx 12.74$ (outside domain) and $x\approx 3.92$ . $V(3.92) = 3.92 \cdot 12.16 \cdot 22.16 \approx 1056$ cm³.
Ex. 70.7
Find the point on the parabola $y = x^2$ closest to the point $(3, 0)$ .
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See the referenced source for the detailed solution.
Ex. 70.8Answer key
A fence of $200\ \text{m}$ is to enclose a rectangular area against a wall (the wall forms one side). Maximize the area.
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Length parallel to the wall: $l$ ; width perpendicular: $w$ . Fencing: $l + 2w = 200$ . Area: $A = lw = (200-2w)w$ . $A'(w)=200-4w=0 \Rightarrow w=50, l=100$ . $A=5000$ m².
Ex. 70.9
Determine the inflection points and concavity of $f(x) = e^{-x^2}$ .
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$f'(x)=-2xe^{-x^2}$ . $f''(x)=-2e^{-x^2}+4x^2e^{-x^2}=2e^{-x^2}(2x^2-1)$ . $f''=0 \Rightarrow x=\pm 1/\sqrt{2}$ . Inflection points at $x=\pm 1/\sqrt{2}\approx \pm 0.707$ . Concave up for $|x|>1/\sqrt{2}$ , concave down for $|x|<1/\sqrt{2}$ .
Ex. 70.10
Perform a complete sketch of $f(x) = \ln(1 + x^2)$ : domain, symmetry, extrema, inflection points, asymptotic behavior.
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Domain: $\mathbb{R}$ . Even function. $f(0)=0$ . $\lim_{|x|\to\infty}f=+\infty$ . $f'(x)=2x/(1+x^2)$ : negative for $x<0$ , zero at $x=0$ (minimum, $f(0)=0$ ), positive for $x>0$ . $f''(x)=(2-2x^2)/(1+x^2)^2$ : inflection points at $x=\pm 1$ ( $f(\pm 1)=\ln 2$ ).
Ex. 70.11Answer key
L'Hôpital's Rule applies directly to $\lim_{x \to 0} \sin x / x$ (form $0/0$ ). Which alternative describes a case where the rule does not directly apply?
Select the correct option
Not applicable — the indeterminacy is not $0/0$ or $\infty/\infty$ Applicable, since $\tan x \to \infty$ as $x \to \pi/2$ Applicable, since both limits are $0$ Applicable, just differentiate the numerator
Select an option first
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$\lim_{x\to 0} \sin x / x$ is indeed $0/0$ and L'Hôpital's Rule applies. But applying L'Hôpital's Rule to $\lim_{x\to 0} x/\sin x$ yields $1/\cos x \to 1$ — the same answer. The "not applicable" option referred to the case $\lim_{x\to\pi/2}\tan x / x = \infty/c$ , which is not an indeterminate form — option A is the most precise for that specific statement.
Ex. 70.12
Calculate $\lim_{x \to 0} \dfrac{1 - \cos x}{x^2}$ .
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Form $0/0$. L'Hôpital's Rule: $\lim_{x\to 0}\sin x / (2x) = \lim \cos x / 2 = 1/2$ .
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See the referenced source for the step-by-step walkthrough.
Ex. 70.13
Calculate $\lim_{x \to +\infty} \dfrac{x^2}{e^x}$ .
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Form $\infty/\infty$. Apply L'Hôpital's Rule twice: $\lim x^2/e^x = \lim 2x/e^x = \lim 2/e^x = 0$. The exponential function grows faster than any power function.
Ex. 70.14
Calculate $\lim_{x \to 0^+} x^{\sin x}$ (indeterminate form $0^0$ ).
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Form $0^0$. Rewrite: $x^{\sin x} = e^{\sin x \ln x}$ . Calculate $\lim_{x\to 0^+}\sin x \ln x = \lim_{x\to 0^+}\ln x / (1/\sin x)$ — form $-\infty/\infty$. L'Hôpital's Rule: $(1/x)/(-\cos x/\sin^2 x) = -\sin^2 x/(x\cos x) \to 0$. Therefore $x^{\sin x} \to e^0 = 1$ .
Ex. 70.15
Write the Maclaurin polynomial of $\sin x$ of order 5 ( $P_5$ ).
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$\sin x = x - x^3/6 + x^5/120 - x^7/5040 + \cdots$. Up to order 5: $P_5(x) = x - x^3/3! + x^5/5! = x - x^3/6 + x^5/120$.
Show step-by-step (with the why)
1. Calculate the derivatives of $\sin x$ at $a=0$: $f(0)=0, f'(0)=1, f''(0)=0, f'''(0)=-1, f^{(4)}(0)=0, f^{(5)}(0)=1$ .
2. Construct the polynomial: $P_5(x) = 0 + x - 0 - x^3/6 + 0 + x^5/120$.
3. Simplify: $P_5(x) = x - x^3/6 + x^5/120$.
Mnemonic: the derivatives of $\sin x$ cycle with period 4: $\sin, \cos, -\sin, -\cos, \sin, \ldots$. Only odd terms survive for $\sin x$, since $\cos(0)=1$ and $\sin(0)=0$.
Ex. 70.16
Use Taylor expansion to calculate $\lim_{x \to 0} \dfrac{\tan x - \sin x}{x^3}$ .
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$\tan x = x + x^3/3 + O(x^5)$ and $\sin x = x - x^3/6 + O(x^5)$. Numerator: $\tan x - \sin x = x^3/3 + x^3/6 + O(x^5) = x^3/2 + O(x^5)$. Limit: $x^3/(2x^3) = 1/2$.
Ex. 70.17Answer key
Approximate $\sqrt{1.1}$ using the Maclaurin series of $(1+x)^{1/2}$ up to order 3.
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$(1+x)^{1/2} = 1 + x/2 - x^2/8 + x^3/16 - \cdots$ (binomial series). For $x=0.1$: $P_3(0.1) = 1 + 0.05 - 0.00125 + 0.0000625 = 1.0488$. Exact value: $\sqrt{1.1} \approx 1.04881$ . Error: $< 10^{-5}$ .
Ex. 70.18
Approximate $e^{-0.1}$ with the Maclaurin polynomial of $e^x$ of order 3. Calculate the error.
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$e^x = 1 + x + x^2/2 + x^3/6 + O(x^4)$ for $x=-0.1$: $P_3(-0.1) = 1 - 0.1 + 0.005 - 0.000167 = 0.904833$. Exact value: $e^{-0.1} \approx 0.904837$ . Error: $4 \times 10^{-6}$ .
Ex. 70.19
Calculate $\lim_{x \to 0} \dfrac{x - \sin x}{x^3}$ using Taylor series.
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$\sin x = x - x^3/6 + x^5/120 - \cdots$. So $\sin x - x = -x^3/6 + O(x^5)$. $(\sin x - x)/x^3 \to -1/6$. But the problem asks for $(x - \sin x)/x^3 \to 1/6$.
Ex. 70.20
Write the Maclaurin series for $\ln(1+x)$ up to the $x^5$ term.
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$\ln(1+x) = x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - \cdots$. Series is valid for $|x| \leq 1, x \neq -1$.
Ex. 70.21
Calculate $\lim_{x \to 0} \dfrac{e^x - \cos x - x}{x^2}$ via Taylor series.
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$e^x = 1 + x + x^2/2 + \cdots$ and $\cos x = 1 - x^2/2 + x^4/24 + \cdots$. Numerator: $e^x - \cos x - x = x^2/2 + x^2/2 + O(x^3) = x^2 + O(x^3)$. Limit: $x^2/x^2 = 1$.
Ex. 70.22
What is the interval of convergence for the Maclaurin series of $\ln(1 + x)$ ?
Select the correct option
The series converges to $f(x)$ for all $\mathbb{R}$ The series is a polynomial and converges everywhere on $\mathbb{R}$ The series converges only for $|x| < 1$ The series converges on $(-1, 1]$ and diverges elsewhere
Select an option first
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The Maclaurin series for $\ln(1+x)$ has a radius of convergence $R=1$. At $x=1$ it converges (alternating harmonic series) and at $x=-1$ it diverges (harmonic series). Thus it converges on $(-1,1]$.
Ex. 70.23
Write the Maclaurin polynomial $P_6(x)$ for $\cos x$ .
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$\cos x = 1 - x^2/2 + x^4/24 - x^6/720 + \cdots$. Up to order 6: $P_6(x) = 1 - x^2/2 + x^4/24 - x^6/720$.
Ex. 70.24
Calculate $\lim_{x \to +\infty} \left(1 + \dfrac{2}{x}\right)^x$ (form $1^\infty$ ).
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Form $1^\infty$. Write $L = \lim (1+2/x)^x$. $\ln L = \lim x \ln(1+2/x)$. Let $t=1/x \to 0^+$: $= \lim \ln(1+2t)/t$ (form $0/0$). L'Hôpital's Rule: $= \lim 2/(1+2t) = 2$. Therefore $L=e^2$.
Ex. 70.25
Cost $C(q) = q^2/4 + 5q + 100$ (Reais), price $p = 50$ Reais/unit. Find the quantity $q^*$ and maximum profit $\pi^*$ .
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$\pi(q) = 50q - (q^2/4 + 5q + 100) = -q^2/4 + 45q - 100$. $\pi'(q) = -q/2 + 45 = 0 \Rightarrow q^*=90$. $\pi(90) = -2025 + 4050 - 100 = 1925$ Reais.
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Ex. 70.26Answer key
A monopolist faces demand $q = 100 - 2p$ and has cost $C(q) = 10q$ . Find $q^*$ , $p^*$ , and the maximum profit.
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See the referenced source for the detailed solution.
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1. Invert the demand function: $q=100-2p \Rightarrow p=(100-q)/2$. Why: to express revenue as a function of $q$.
2. Revenue: $R(q)=pq=q(100-q)/2=-q^2/2+50q$.
3. Profit: $\pi(q)=R-C=-q^2/2+50q-10q=-q^2/2+40q$.
4. Condition for maximum: $\pi'=-q+40=0 \Rightarrow q^*=40$. Confirm: $\pi''=-1<0$ .
5. Optimal price: $p^*=(100-40)/2=30$. Profit: $\pi(40)=-800+1600=800$.
Mnemonic: the monopolist's condition is $MR=MC$: $MR=R'(q)=-q+50$ and $MC=C'(q)=10$, so $-q+50=10 \Rightarrow q=40$.
Ex. 70.27
A ball is thrown vertically upward with an initial velocity $v_0 = 20\ \text{m/s}$ ( $g = 10\ \text{m/s}^2$ ). Calculate the maximum height and the time of flight.
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$s(t)=20t-5t^2$ (with $g=10$ m/s²). $v(t)=20-10t=0 \Rightarrow t=2$ s. Max height: $s(2)=40-20=20$ m. Ground: $s(t)=0 \Rightarrow t(20-5t)=0 \Rightarrow t=4$ s.
Ex. 70.28Answer key
Given $s(t) = t^3 - 9t^2 + 24t$ . When is $v(t) = 0$ ? Calculate the total distance traveled over $0 \leq t \leq 5$ .
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$v(t) = 3t^2 - 18t + 24 = 3(t-2)(t-4)$. $v=0$ at $t=2$ and $t=4$. Total distance: $|s(2)-s(0)|+|s(4)-s(2)|+|s(5)-s(4)|$. $s(0)=0, s(2)=8-36+48=20, s(4)=64-144+96=16, s(5)=125-225+120=20$. Distance: $20+4+4=28$ m.
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Ex. 70.29
Mass-spring system: $m = 0.5\ \text{kg}$ , $k = 50\ \text{N/m}$ , amplitude $A = 0.1\ \text{m}$ . Calculate the period and the maximum velocity.
Solve online APEX Calculus · §2.2
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Period: $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.5/50} = 2\pi \cdot 0.1 = 0.628$ s. Maximum velocity: $v_{\max} = A\omega = A\sqrt{k/m} = 0.1\sqrt{100} = 1$ m/s.
Ex. 70.30
Use Newton-Raphson on $f(x) = x^2 - 7$ with $x_0 = 3$ to calculate $\sqrt{7}$ to 5 decimal places.
Solve online APEX Calculus · §6.1
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Ex. 70.31
Use Newton-Raphson on $f(x) = x^3 - 2$ with $x_0 = 1$ to approximate $\sqrt[3]{2}$ . Perform 3 iterations.
Solve online APEX Calculus · §6.1
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$f(x)=x^3-2$, $f'(x)=3x^2$, $x_0=1$. $x_1=1-(1-2)/3=1+1/3=1.333$. $x_2=1.333-(1.333^3-2)/(3\cdot 1.333^2)\approx 1.264$. $x_3\approx 1.25993$. Exact value: $2^{1/3}\approx 1.25992$ .
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Ex. 70.32
Apply Newton-Raphson to $f(x) = e^x - 3x$ to find both real roots. Use $x_0 = 0$ for one and $x_0 = 1.5$ for the other.
Solve online APEX Calculus · §6.1
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Roots of $e^x=3x$: one in $(0,1)$ (approx. $x_1\approx 0.619$) and another in $(1,2)$ (approx. $x_2\approx 1.512$). For each root, use a nearby $x_0$ and iterate.
Ex. 70.33
Kepler's equation is $E - 0.3\sin E = 1$ . Use Newton-Raphson with $E_0 = 1$ and perform 4 iterations.
Solve online OpenStax Calculus Vol. 1 · §4.9
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Kepler's equation: $E - 0.3\sin E = 1$. Rewrite: $f(E)=E-0.3\sin E - 1 = 0$. $f'(E)=1-0.3\cos E$. For $E_0=1$: $E_1=1-(1-0.3\sin 1-1)/(1-0.3\cos 1)=1-(-0.252)/(0.838)\approx 1.301$. $E_2\approx 1.351$. $E_3\approx 1.352$. $E_4\approx 1.352$ (converged).
Ex. 70.34
Show that a strictly convex ( $f'' > 0$ ) $C^2$ function on $\mathbb{R}$ has at most one minimum point.
Solve online Active Calculus · §3.3
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$f''\geq 0$ on all $\mathbb{R}$ implies $f$ is convex. If $f$ had two local minima $a$ and $b$, by the Mean Value Theorem there would exist $c \in (a,b)$ with $f(c) < f(a)$ and $f(c) < f(b)$ . But convexity requires $f(c) \leq f(a) + (f(b)-f(a))/(b-a)\cdot(c-a)$. For a strictly convex function ( $f''>0$ ), only one global minimum exists.
Ex. 70.35Answer key
Use the Maclaurin series of $\sin x$ to prove that $\sin x < x$ for all $x > 0$ .
Solve online Active Calculus · §8.4
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$\sin x = x - x^3/6 + \cdots$. For $x>0$ : $x - \sin x = x^3/6 - x^5/120 + \cdots > 0$ since the first positive term dominates. Therefore $\sin x < x$ for all $x>0$ .
Ex. 70.36
Sketch $f(x) = x^x$ on $(0, +\infty)$ . Find the minimum and analyze the behavior at the domain's extremes.
Solve online APEX Calculus · §3.5
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Ex. 70.37
Newton-Raphson applied to $f(x) = \arctan x$ with $x_0 = 2$ diverges. Explain geometrically why, and show numerically.
Solve online Active Calculus · §3.4
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Ex. 70.38Answer key
Prove via Taylor series: if $f'(a) = 0$ and $f^{(k)}(a)$ is the first non-zero derivative at $a$ , then $a$ is an extremum if $k$ is even, and a saddle point/inflection point if $k$ is odd.
Active Calculus · §3.1
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Taylor expansion around $a$: $f(x) = f(a) + f'(a)(x-a) + f^{(k)}(a)/k! \cdot (x-a)^k + O((x-a)^{k+1})$ with $f'(a)=\cdots=f^{(k-1)}(a)=0$ and $f^{(k)}(a)\neq 0$ . For $x$ near $a$, the sign of $f(x)-f(a)$ is determined by $f^{(k)}(a)(x-a)^k/k!$ . If $k$ is even: $(x-a)^k > 0$ for $x \neq a$, so $f(x)-f(a)$ has constant sign — an extremum. If $k$ is odd: $(x-a)^k$ changes sign — not an extremum (inflection point).
Ex. 70.39
Show, via Taylor of order 1, that $\lim_{x\to a} f(x)/g(x) = f'(a)/g'(a)$ when $f(a) = g(a) = 0$ and $g'(a) \neq 0$ .
Solve online Active Calculus · §2.6
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Substitute $f(x)\approx f'(a)(x-a)$ and $g(x)\approx g'(a)(x-a)$ (Taylor of order 1, with $f(a)=g(a)=0$). Then $f(x)/g(x) \approx f'(a)(x-a)/(g'(a)(x-a)) = f'(a)/g'(a)$ (assuming $g'(a)\neq 0$). This "proves" L'Hôpital's Rule as a corollary of Taylor of order 1 — the rigorous proof uses Cauchy's MVT.
Ex. 70.40Answer key
Formally derive the $MR = MC$ condition for maximum profit. Explain why a monopolist produces less than a competitive firm.
Solve online Active Calculus · §3.4
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Sources

Active Calculus — Matt Boelkins, David Austin, Steve Schlicker · Grand Valley State University · 2024 · CC-BY-NC-SA. Sections §2.6 (L'Hôpital's Rule), §3.1–3.4 (optimization), §8.4–8.5 (Taylor series).
APEX Calculus — Gregory Hartman et al. · Virginia Military Institute · 2024 · CC-BY-NC. Chapters 3 (function analysis), 4 (applications), 6 (Newton's Method and applications), 8 (Taylor series).
Calculus Volume 1 — OpenStax (Strang, Herman et al.) · 2023 · CC-BY-NC-SA. Sections §4.3 (max-min), §4.7 (applied optimization), §4.8 (L'Hôpital's Rule), §4.9 (Newton's Method).