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Extreme values, the Mean Value Theorem and the shape of a graph
Absolute and local extrema, Rolle and the MVT, monotonicity, concavity, asymptotes and curve sketching.
Lecture 5 · Tutorial 5 · Applications of derivatives and antiderivatives
Textbook sections: Thomas ch. 4 Applications of Derivatives (4.1–4.8) · Stewart ch. 4 Applications of Differentiation (4.1–4.9)
Textbook sections: Thomas ch. 4 Applications of Derivatives (4.1–4.8) · Stewart ch. 4 Applications of Differentiation (4.1–4.9)
Extreme values, the Mean Value Theorem and the shape of a graph
Calculus I · Module 6
After this module you can
- Find absolute extrema on a closed interval and local extrema with the first and second derivative tests.
- Apply Rolle's theorem and the MVT correctly (check the hypotheses).
- Find inflection points and asymptotes, and sketch a curve.
Extreme values
- Extreme Value Theorem: $f$ continuous on $[a,b]$ has an absolute max and min.
- Fermat: a local extremum at an interior point where $f'$ exists has $f'(c)=0$ (the converse is false: $x^3$).
- Closed-interval method: evaluate $f$ at critical points ($f'=0$ or undefined) and at the endpoints; compare.
- Example: $x^2-1$ on $[-1,2]$: values $0,-1,3$: max $3$, min $-1$.
Rolle and the Mean Value Theorem
- Rolle: continuous on $[a,b]$, differentiable on $(a,b)$, $f(a)=f(b)$ $\Rightarrow$ $f'(c)=0$ somewhere.
- MVT: same hypotheses, no condition on endpoints $\Rightarrow$ $f'(c)=\frac{f(b)-f(a)}{b-a}$.
- A thermometer rising $-19\to100$ °C in $14$ s was rising at exactly $8.5$ °C/s at some instant.
- $\sqrt{x(1-x)}$ satisfies the MVT on $[0,1]$: differentiability is only needed on the open interval.
Increasing, decreasing, concavity
| Test | Conclusion |
|---|---|
| $f'>0$ on an interval | increasing |
| $f'$ changes $+\to-$ at $c$ | local maximum |
| $f'$ changes $-\to+$ at $c$ | local minimum |
| $f''>0$ | concave up |
| $f''$ changes sign at $c$ | inflection point |
| $f'(c)=0$, $f''(c)>0$ ($<0$) | local min (max) |
Asymptotes and sketching
- Vertical: $\lim f=\pm\infty$ at $x=a$. Horizontal: $\lim_{x\to\pm\infty}f=L$.
- Oblique: $f=mx+c+\text{(vanishing)}$; for rational $f$ use long division: $\frac{2x^2+3}{x-1}=2x+2+\frac{5}{x-1}$.
- Sketch: domain, intercepts, symmetry, asymptotes, $f'$ (monotonic, extrema), $f''$ (concavity, inflection).
- $y=x+\sin2x$ on $[-\frac{2\pi}3,\frac{2\pi}3]$: local min at $-\frac\pi3$, local max at $\frac\pi3$, inflections at $0,\pm\frac\pi2$.
Typical mistakes
EndpointsAbsolute extrema can sit at endpoints: always evaluate them.
$f'=0$ is not enoughCheck the sign change (or $f''$).
Undefined $f'$A critical point can be where $f'$ does not exist (e.g. $x^{2/3}$ at $0$).
Practice next
- Chapter quizzes Ch. 3 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Curve-sketching helper
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · First derivative test (Computation problems 5, Q7)
$h(x)=\sqrt[3]{x}\,(x^2-4)$ has critical points where $h'(x)=\frac{7x^2-4}{3x^{2/3}}$ is $0$ or undefined, namely
Show the answer and the solution
Answer: $x=0$ (undefined) and $x=\pm\frac{2}{\sqrt7}$; the local extrema are only at $\pm\frac{2}{\sqrt7}$
$h=x^{7/3}-4x^{1/3}$, $h'=\frac73x^{4/3}-\frac43x^{-2/3}=\frac{7x^2-4}{3x^{2/3}}$. At $x=0$, $h'$ is undefined but $h'$ does not change sign there (the factor $x^{2/3}>0$), so no extremum.
Worked example 2 · Concavity and inflection (Computation problems 5, Q10)
For $y=x+\sin2x$ on $[-\frac{2\pi}{3},\frac{2\pi}{3}]$ the inflection points are at
Show the answer and the solution
Answer: $x=0$ and $x=\pm\frac{\pi}{2}$
$y''=-4\sin2x=0\Rightarrow2x=k\pi\Rightarrow x=0,\pm\frac\pi2$ (the points $x=\pm\frac\pi3$ are the local extrema, where $y'=1+2\cos2x=0$).
Worked example 3 · Mean Value Theorem (Tutorial 5, Q6(a))
To show $f(x)=x^4+3x+1$ has exactly one zero in $[-2,-1]$ one uses
Show the answer and the solution
Answer: the IVT ($f(-2)=11>0$, $f(-1)=-1<0$) for existence and $f'(x)=4x^3+3<0$ on $[-2,-1]$ (Rolle) for uniqueness
A sign change gives a zero; if there were two zeros, Rolle would give a zero of $f'$ between them, but $f'<0$ throughout.
Worked example 4 · Extreme values (Computation problems 5, Q2)
Find the absolute maximum of $g(x)=\sqrt{4-x^2}$ on $[-2,1]$.
Show the answer and the solution
Answer: $2$
$g'(x)=-\frac{x}{\sqrt{4-x^2}}=0$ at $x=0$ where $g=2$; endpoints: $g(-2)=0$, $g(1)=\sqrt3$. Maximum $=2$.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Curve-sketching helper