13b. Law of the Mean — Practice

Drill sheet for 13. Law of the Mean. Pattern matches the 初二 worksheets: book example above, then EquationWorksheet batches (show/hide solution · regenerate). Proofs use guided choice.

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1. Rolle — find $x_0$

Book. $f(x)=x^{3}-12x$ on $[0,\,2\sqrt{3}]$. Endpoints are zeros; $f'(x)=3x^{2}-12=0\Rightarrow x=\pm 2$; take $x_0=2$.

Rolle: find x₀

Cubics x³ − 3a²x on [0, a√3] — endpoint zeros, then f′=0

(1)Find the value of $x_0$ prescribed by Rolle’s Theorem for $f(x)=x^{3}-12x$ on $[0,\,2\sqrt{3}]$. Note $f(0)=f(2\sqrt{3})=0$.

Compute $f'(x)=3x^{2}-12$.

Set $f'(x)=0$: $3x^{2}=12\Rightarrow x=\pm 2$.

Only $x=2$ lies in $(0,\,2\sqrt{3})$.

Therefore $x_0=2$.

(2)Find $x_0$ from Rolle’s Theorem for $f(x)=x^{3}-3x$ on $[0,\,1\sqrt{3}]$. (Check the endpoints are zeros first.)

$f(0)=0$ and $f(1\sqrt{3})=(1\sqrt{3})^{3}-3(1\sqrt{3})=0$.

$f'(x)=3x^{2}-3=3(x^{2}-1)$.

$f'(x)=0\Rightarrow x=\pm 1$. In $(0,\,1\sqrt{3})$ we take $x_0=1$.

$x_0=1$.

(3)Find $x_0$ from Rolle’s Theorem for $f(x)=x^{3}-48x$ on $[0,\,4\sqrt{3}]$. (Check the endpoints are zeros first.)

$f(0)=0$ and $f(4\sqrt{3})=(4\sqrt{3})^{3}-48(4\sqrt{3})=0$.

$f'(x)=3x^{2}-48=3(x^{2}-16)$.

$f'(x)=0\Rightarrow x=\pm 4$. In $(0,\,4\sqrt{3})$ we take $x_0=4$.

$x_0=4$.

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2. Does Rolle apply?

Book. On $[0,4]$: (a) $\dfrac{x^{2}-4x}{x-2}$ fails (discontinuity at $2$). (b) $\dfrac{x^{2}-4x}{x+2}$ works (pole at $-2$ outside); $x_0=2(\sqrt{3}-1)$.

Rolle: apply or not?

Check continuity on [a,b] before hunting for f′=0

(1)Does Rolle’s Theorem apply to $f(x)=\dfrac{x(x-6)}{x-3}$ on $[0,6]$? If not, say why.

$f(0)=f(6)=0$, but $f$ has a discontinuity at $x=3$ inside $[0,6]$.

Continuity on the closed interval fails, so Rolle’s Theorem does **not** apply.

Does not apply (discontinuity inside the interval).

(2)Does Rolle’s Theorem apply to $f(x)=\dfrac{x(x-6)}{x-3}$ on $[0,6]$? If not, say why.

$f(0)=f(6)=0$, but $f$ has a discontinuity at $x=3$ inside $[0,6]$.

Continuity on the closed interval fails, so Rolle’s Theorem does **not** apply.

Does not apply (discontinuity inside the interval).

(3)Does Rolle’s Theorem apply to $f(x)=\dfrac{x(x-8)}{x-(-3)}$ on $[0,8]$?

$f(0)=f(8)=0$. The pole $x=-3$ is **outside** $[0,8]$.

On $[0,8]$ the function is continuous, and differentiable on $(0,8)$.

So Rolle’s Theorem applies.

Yes — Rolle’s Theorem applies.

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3. Law of the Mean — quadratic

Book. $f(x)=3x^{2}+4x-3$, $a=1$, $b=3$ → secant $16$, $6x_0+4=16$ → $x_0=2$ (midpoint).

MVT for a quadratic

Secant slope = f′(x₀); for ax²+bx+c the point is (a+b)/2

(1)Find $x_0$ from the Law of the Mean for $f(x)=2x^{2}+2x+2$ on $[2,5]$.

$f(2)=14$, $f(5)=62$, secant slope $=\dfrac{62-(14)}{5-2}=16$.

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

Solve 4x_0+2=16$. For a quadratic,$x_0=\dfrac{a+b}{2}$.

$x_0=3.5$.

(2)Find $x_0$ from the Law of the Mean for $f(x)=2x^{2}-2x-5$ on $[-1,3]$.

$f(-1)=-1$, $f(3)=7$, secant slope $=\dfrac{7-(-1)}{3--1}=2$.

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

Solve 4x_0-2=2$. For a quadratic,$x_0=\dfrac{a+b}{2}$.

$x_0=1$.

(3)Find $x_0$ from the Law of the Mean for $f(x)=4x^{2}-4x-5$ on $[2,5]$.

$f(2)=3$, $f(5)=75$, secant slope $=\dfrac{75-(3)}{5-2}=24$.

$f'(x)=8x-4$.

Solve 8x_0-4=24$. For a quadratic,$x_0=\dfrac{a+b}{2}$.

$x_0=3.5$.

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4. Extended Law of the Mean

Book. $f(x)=3x+2$, $g(x)=x^{2}+1$ on $[1,4]$ → $\dfrac{3}{5}=\dfrac{3}{2x_0}$ → $x_0=\dfrac{5}{2}$.

Extended MVT

Linear f, quadratic g — equate (Δf/Δg) = f′/g′

(1)Find $x_0$ from the extended Law of the Mean for $f(x)=4x-2$ and $g(x)=x^{2}+2$ on $[1,3]$.

$\dfrac{f(3)-f(1)}{g(3)-g(1)}=\dfrac{10-(2)}{11-3}=\dfrac{1}{1}$.

$\dfrac{f'}{g'}=\dfrac{4}{2x_0}$.

Equate and solve: $x_0=2$.

$x_0=2$.

(2)Find $x_0$ from the extended Law of the Mean for $f(x)=3x-1$ and $g(x)=x^{2}$ on $[1,3]$.

$\dfrac{f(3)-f(1)}{g(3)-g(1)}=\dfrac{8-(2)}{9-1}=\dfrac{3}{4}$.

$\dfrac{f'}{g'}=\dfrac{3}{2x_0}$.

Equate and solve: $x_0=2$.

$x_0=2$.

(3)Find $x_0$ from the extended Law of the Mean for $f(x)=4x-1$ and $g(x)=x^{2}+1$ on $[2,5]$.

$\dfrac{f(5)-f(2)}{g(5)-g(2)}=\dfrac{19-(7)}{26-5}=\dfrac{4}{7}$.

$\dfrac{f'}{g'}=\dfrac{4}{2x_0}$.

Equate and solve: $x_0=\dfrac{7}{2}$.

$x_0=\dfrac{7}{2}$.

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5–7. Proofs (guided)

5. Relative extremum ⇒ $f'=0$

6. Rolle’s Theorem

7. Law of the Mean

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8. Negating monotone

Book. If $g$ is increasing and $u<v$, then $g(u)<g(v)$, so $-g(u)>-g(v)$: $-g$ is decreasing.

Quiz

Negating an increasing function

1 / 1

If g is increasing on an interval, then −g is…

Pick an answer to see if you are right.

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9. Prove Theorem 13.7 (monotone from $f'$)

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10–12. Increasing / decreasing / unique root

Book 10. $f(x)=x^{5}+20x-6$, $f'=5x^{4}+20>0$.

Book 11. $f(x)=1-x^{3}-x^{7}$, $f'\\le 0$ (strict off $0$), plus comparison across $0$.

Book 12. $4x^{3}+x-3=0$: IVT on $(0,1)$ and $f'=12x^{2}+1>0$ ⇒ unique root.

Show increasing

Produce f′>0 everywhere, cite Theorem 13.7(a)

(1)Show that $f(x)=x^{5}+12x$ is increasing for all real $x$.

$f'(x)=5x^{4}+12$.

Since 4 is even, $x^{4}\ge 0$, so $f'(x)\ge 12>0$.

Theorem 13.7(a) $\Rightarrow$ $f$ increasing on $\mathbb{R}$.

$f$ is increasing everywhere.

(2)Show that $f(x)=x^{7}+20x$ is increasing for all real $x$.

$f'(x)=7x^{6}+20$.

Since 6 is even, $x^{6}\ge 0$, so $f'(x)\ge 20>0$.

Theorem 13.7(a) $\Rightarrow$ $f$ increasing on $\mathbb{R}$.

$f$ is increasing everywhere.

Show decreasing

Produce f′<0 (or ≤0 with a zero check), cite Theorem 13.7(b)

(1)Show that $f(x)=-x^{5}-7x$ is decreasing for all real $x$.

$f'(x)=-5x^{4}-7$.

Even power $x^{4}\ge 0\Rightarrow f'(x)\le -7<0$.

Theorem 13.7(b) $\Rightarrow$ decreasing on $\mathbb{R}$.

$f$ is decreasing everywhere.

(2)Show that $f(x)=-x^{3}-5x$ is decreasing for all real $x$.

$f'(x)=-3x^{2}-5$.

Even power $x^{2}\ge 0\Rightarrow f'(x)\le -5<0$.

Theorem 13.7(b) $\Rightarrow$ decreasing on $\mathbb{R}$.

$f$ is decreasing everywhere.

Exactly one real root

IVT for existence + f′>0 for uniqueness

(1)Show that $f(x)=4x^{3}+x-3=0$ has exactly one real solution.

$f(0)=-3<0$ and $f(1)=2>0$, so IVT gives a root in $(0,1)$.

$f'(x)=12x^{2}+1>0$ for all $x$, so $f$ is strictly increasing.

A strictly increasing continuous function crosses zero at most once.

Exactly one real root (in $(0,1)$).

(2)Show that $f(x)=4x^{3}+x-3=0$ has exactly one real solution.

$f(0)=-3<0$ and $f(1)=2>0$, so IVT gives a root in $(0,1)$.

$f'(x)=12x^{2}+1>0$ for all $x$, so $f$ is strictly increasing.

A strictly increasing continuous function crosses zero at most once.

Exactly one real root (in $(0,1)$).

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13. Extended MVT — proof sketch

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14. Higher-order Law of the Mean (finish)

After defining $K$ so identity (2) holds and building the auxiliary $F$, Rolle (applied carefully) produces a point where $F'=0$. Differentiating $F$ makes intermediate terms telescope:

$$\begin{aligned} F'(x_0)=& f'(x_0)+\bigl[f''(x_0)(b-x_0)-f'(x_0)\bigr] +\Bigl[\frac{f'''(x_0)}{2!}(b-x_0)^{2}-f''(x_0)(b-x_0)\Bigr]\\ &+\cdots+\Bigl[\frac{f^{(n)}(x_0)}{(n-1)!}(b-x_0)^{n-1}-\frac{f^{(n-1)}(x_0)}{(n-2)!}(b-x_0)^{n-2}\Bigr] -Kn(b-x_0)^{n-1}\\ =& \frac{f^{(n)}(x_0)}{(n-1)!}(b-x_0)^{n-1}-Kn(b-x_0)^{n-1}=0. \end{aligned}$$

Hence $K=\dfrac{f^{(n)}(x_0)}{n!}$, and (2) becomes the higher-order mean-value formula (1).

Quiz

Higher-order mean value

1 / 1

The higher-order Law of the Mean is essentially…

Pick an answer to see if you are right.

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15. Vanishing derivative ⇒ constant

Claim. If $f'=0$ on $(a,b)$, then $f$ is constant on $(a,b)$.

Idea. For $u<v$ in $(a,b)$, MVT gives $\dfrac{f(v)-f(u)}{v-u}=f'(x_0)=0$, so $f(v)=f(u)$.

Problem 18

If $f'=g'$ on $(a,b)$, then $(f-g)'=0$, so Problem 15 yields $f=g+K$.

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Mixed review (core)

Law of the Mean — mixed set

Rolle / MVT / extended / monotone / unique root

(1)Does Rolle’s Theorem apply to $f(x)=\dfrac{x(x-4)}{x-2}$ on $[0,4]$? If not, say why.

$f(0)=f(4)=0$, but $f$ has a discontinuity at $x=2$ inside $[0,4]$.

Continuity on the closed interval fails, so Rolle’s Theorem does **not** apply.

Does not apply (discontinuity inside the interval).

(2)Show that $f(x)=4x^{3}+x-3=0$ has exactly one real solution.

$f(0)=-3<0$ and $f(1)=2>0$, so IVT gives a root in $(0,1)$.

$f'(x)=12x^{2}+1>0$ for all $x$, so $f$ is strictly increasing.

A strictly increasing continuous function crosses zero at most once.

Exactly one real root (in $(0,1)$).

(3)Find $x_0$ from the extended Law of the Mean for $f(x)=3x-2$ and $g(x)=x^{2}+2$ on $[1,3]$.

$\dfrac{f(3)-f(1)}{g(3)-g(1)}=\dfrac{7-(1)}{11-3}=\dfrac{3}{4}$.

$\dfrac{f'}{g'}=\dfrac{3}{2x_0}$.

Equate and solve: $x_0=2$.

$x_0=2$.

(4)Show that $f(x)=-x^{5}-7x$ is decreasing for all real $x$.

$f'(x)=-5x^{4}-7$.

Even power $x^{4}\ge 0\Rightarrow f'(x)\le -7<0$.

Theorem 13.7(b) $\Rightarrow$ decreasing on $\mathbb{R}$.

$f$ is decreasing everywhere.

(5)Find $x_0$ from the Law of the Mean for $f(x)=2x^{2}+2$ on $[2,5]$.

$f(2)=10$, $f(5)=52$, secant slope $=\dfrac{52-(10)}{5-2}=14$.

$f'(x)=4x$.

Solve 4x_0=14$. For a quadratic,$x_0=\dfrac{a+b}{2}$.

$x_0=3.5$.

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Bonus · Supplementary problems

16–17. Rolle / MVT spot checks

Book 16. $f(x)=x^{2}-4x+3$ on $[1,3]$ → $x_0=2$.

Book 17. (a) $y=x^{3}$ on $[0,6]$ → $x_0=2\sqrt{3}$. (b) Any quadratic on $[x_1,x_2]$ → midpoint.

Supp: Rolle on a quadratic

Endpoint zeros → solve f′=0

(1)If $f(x)=x^{2}-4x+3$ on $[1,3]$, find a value prescribed by Rolle’s Theorem.

$f(1)=0=f(3)$.

$f'(x)=2x-4=0\Rightarrow x=2\in(1,3)$.

$x_0=2$.

(2)If $f(x)=x^{2}-4x+3$ on $[1,3]$, find $x_0$ from Rolle’s Theorem.

$f(1)=f(3)=0$.

$f'(x)=2x-4=0\Rightarrow x_0=2$.

$x_0=2$.

(3)If $f(x)=x^{2}-9x+18$ on $[3,6]$, find $x_0$ from Rolle’s Theorem.

$f(3)=f(6)=0$.

$f'(x)=2x-9=0\Rightarrow x_0=4.5$.

$x_0=4.5$.

Supp: MVT (cube or quadratic)

y=x³ on [0,b], or midpoint for ax²+bx+c

(1)Find a value prescribed by the Law of the Mean for $y=x^{3}$ on $[0,6]$.

Secant: $\dfrac{6^{3}-0}{6-0}=36$.

$y'=3x^{2}$. Set $3x_0^{2}=36\Rightarrow x_0^{2}=12\Rightarrow x_0=2\sqrt{3}$ (positive).

$x_0=2\sqrt{3}$.

(2)Find $x_0$ from the Law of the Mean for $y=x^{3}$ on $[0,9]$.

Secant $=\dfrac{9^{3}}{9}=81$.

$3x_0^{2}=81\Rightarrow x_0=\dfrac{9}{\sqrt{3}}$ (in $(0,9)$).

$x_0=\dfrac{9}{\sqrt{3}}$.

(3)Find $x_0$ from the Law of the Mean for $y=x^{3}$ on $[0,3]$.

Secant $=\dfrac{3^{3}}{3}=9$.

$3x_0^{2}=9\Rightarrow x_0=\dfrac{3}{\sqrt{3}}$ (in $(0,3)$).

$x_0=\dfrac{3}{\sqrt{3}}$.

19. Extended MVT (numbers)

Book. $f(x)=x^{2}+2x-3$, $g(x)=x^{2}-4x+6$ on $[0,1]$ → $x_0=\dfrac{1}{2}$.

Supp: extended MVT

Equate Δf/Δg = f′/g′

(1)Find $x_0$ from the extended Law of the Mean for $f(x)=3x+2$ and $g(x)=x^{2}+1$ on $[1,4]$.

$\dfrac{f(4)-f(1)}{g(4)-g(1)}=\dfrac{14-5}{17-2}=\dfrac{3}{5}$.

$\dfrac{f'(x_0)}{g'(x_0)}=\dfrac{3}{2x_0}$.

So $\dfrac{3}{2x_0}=\dfrac{3}{5}\Rightarrow x_0=\dfrac{5}{2}$.

$x_0=\dfrac{5}{2}$.

(2)Find $x_0$ from the extended Law of the Mean for $f(x)=2x$ and $g(x)=x^{2}+1$ on $[2,4]$.

$\dfrac{f(4)-f(2)}{g(4)-g(2)}=\dfrac{8-(4)}{17-5}=\dfrac{1}{3}$.

$\dfrac{f'}{g'}=\dfrac{2}{2x_0}$.

Equate and solve: $x_0=3$.

$x_0=3$.

20. Cubic $x^{3}+px+q=0$

Quiz

Depressed cubic root count

1 / 2

For x³ + p x + q = 0 with p > 0, why is there exactly one real root?

Pick an answer to see if you are right.

21. Linear fractional — no relative extrema

22. Unique real root

Book. $5x^{3}+11x-20=0$ — IVT + $f'>0$.

Supp: unique cubic root

Sign change + strictly positive derivative

(1)Show that $f(x)=5x^{3}+11x-20=0$ has exactly one real solution.

$f(1)=5+11-20=-4<0$, $f(2)=40+22-20=42>0$ $\Rightarrow$ a root in $(1,2)$.

$f'(x)=15x^{2}+11>0$ $\Rightarrow$ strictly increasing $\Rightarrow$ unique real root.

Exactly one real root.

(2)Show that $f(x)=4x^{3}+x-2=0$ has exactly one real root.

$f(0)=-2<0$, $f(1)=3>0$ $\Rightarrow$ a root in $(0,1)$ by IVT.

$f'(x)=12x^{2}+1>0$ $\Rightarrow$ $f$ strictly increasing $\Rightarrow$ at most one root.

Exactly one real root.

23. Increasing / decreasing intervals

Book samples: linear; $x^{2}+6x-11$; $5+8x-x^{2}$; $\sqrt{4-x^{2}}$; $|x-2|+3$; $\dfrac{x}{x^{2}-4}$.

Supp: mono intervals

Sign-chart f′ (watch domain and corners)

(1)Where is $f(x)=3x+5$ increasing / decreasing?

$f'(x)=3>0$.

Increasing on $\mathbb{R}$; never decreasing.

(2)Where is $f(x)=x^{2}+6x-11$ increasing / decreasing?

$f'(x)=2x+6=2(x+3)$.

Sign chart: $f'<0$ on $(-\infty,-3)$, $f'>0$ on $(-3,\infty)$.

Decreasing on $(-\infty,-3)$; increasing on $(-3,+\infty)$.

(3)Where is $f(x)=\dfrac{x}{x^{2}-4}$ increasing / decreasing?

Domain: $x\ne\pm 2$.

$f'(x)=\dfrac{-(x^{2}+4)}{(x^{2}-4)^{2}}<0$ wherever defined.

Decreasing on $(-\infty,-2)$, $(-2,2)$, and $(2,+\infty)$; never increasing.

(4)Where is $f(x)=\dfrac{x}{x^{2}-4}$ increasing / decreasing?

Domain: $x\ne\pm 2$.

$f'(x)=\dfrac{-(x^{2}+4)}{(x^{2}-4)^{2}}<0$ wherever defined.

Decreasing on $(-\infty,-2)$, $(-2,2)$, and $(2,+\infty)$; never increasing.

Graphing check (GC)

For calculator labs (supp 23b / 24): plot $f$ and $f'$, or use a zero finder on $f'$. Example: $f(x)=x^{5}+2x^{3}-6x+1$ — estimate where $f'>0$ vs $f'<0$ from the graph of $f'(x)=5x^{4}+6x^{2}-6$.

25. Rolle checklist

Book. Absolute values / fractional powers often kill differentiability; unequal endpoints kill Rolle even if the graph “looks peaked.”

Supp: does Rolle apply?

Endpoints · continuity · differentiability · then find x₀

(1)Does Rolle apply to $f(x)=\dfrac{x^{2}-3x-4}{x-5}$ on $[-1,4]$? If yes, find $x_0$.

Pole at $x=5$ is outside $[-1,4]$; $f(-1)=f(4)=0$.

Simplify / differentiate: Rolle applies.

Solving $f'(x_0)=0$ yields $x_0=5-\sqrt{6}$ (the root in $(-1,4)$).

Yes — $x_0=5-\sqrt{6}$.

(2)Does Rolle apply to $f(x)=|x^{2}-4|$ on $[0,1]$?

$f(0)=4$, $f(1)=3$, so $f(0)\ne f(1)$.

Equal endpoint values fail.

No — $f(0)\ne f(1)$.

(3)Does Rolle’s Theorem apply to $f(x)=\dfrac{x(x-8)}{x-4}$ on $[0,8]$? If not, say why.

$f(0)=f(8)=0$, but $f$ has a discontinuity at $x=4$ inside $[0,8]$.

Continuity on the closed interval fails, so Rolle’s Theorem does **not** apply.

Does not apply (discontinuity inside the interval).

(4)Does Rolle apply to $f(x)=|x^{2}-4|$ on $[0,1]$?

$f(0)=4$, $f(1)=3$, so $f(0)\ne f(1)$.

Equal endpoint values fail.

No — $f(0)\ne f(1)$.

Bonus mixed

Supplementary mixed set

Problems 16–25 style

(1)If $f(x)=x^{2}-10x+21$ on $[3,7]$, find $x_0$ from Rolle’s Theorem.

$f(3)=f(7)=0$.

$f'(x)=2x-10=0\Rightarrow x_0=5$.

$x_0=5$.

(2)If $f(x)=x^{2}-4x+3$ on $[1,3]$, find a value prescribed by Rolle’s Theorem.

$f(1)=0=f(3)$.

$f'(x)=2x-4=0\Rightarrow x=2\in(1,3)$.

$x_0=2$.

(3)Find $x_0$ from the extended Law of the Mean for $f(x)=3x+2$ and $g(x)=x^{2}+1$ on $[1,3]$.

$\dfrac{f(3)-f(1)}{g(3)-g(1)}=\dfrac{11-(5)}{10-2}=\dfrac{3}{4}$.

$\dfrac{f'}{g'}=\dfrac{3}{2x_0}$.

Equate and solve: $x_0=2$.

$x_0=2$.

(4)Find $x_0$ from the extended Law of the Mean for $f(x)=3x$ and $g(x)=x^{2}+2$ on $[1,4]$.

$\dfrac{f(4)-f(1)}{g(4)-g(1)}=\dfrac{12-(3)}{18-3}=\dfrac{3}{5}$.

$\dfrac{f'}{g'}=\dfrac{3}{2x_0}$.

Equate and solve: $x_0=\dfrac{5}{2}$.

$x_0=\dfrac{5}{2}$.

(5)Does Rolle’s Theorem apply to $f(x)=\dfrac{x(x-8)}{x-(-3)}$ on $[0,8]$?

$f(0)=f(8)=0$. The pole $x=-3$ is **outside** $[0,8]$.

On $[0,8]$ the function is continuous, and differentiable on $(0,8)$.

So Rolle’s Theorem applies.

Yes — Rolle’s Theorem applies.

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Self-check

Quiz

Law of the Mean practice review

1 / 4

Rolle needs f(a)=f(b)=0 (or equal values). What else is essential?

Pick an answer to see if you are right.

Further Reading

• 14. Maximum and Minimum Values — next: classify extrema and absolute max/min on intervals
• 14b. Max/Min — Practice — extrema + applied optimization drills
• 13. Law of the Mean · Increasing and Decreasing — theory and explorers