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Q. 57

Expert-verifiedFound in: Page 249

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Determine whether or not each function f in Exercises 53–60 satisfies the hypotheses of the Mean Value Theorem on the given interval [a, b]. For those that do, use derivatives and algebra to find the exact values of all c ∈ (a, b) that satisfy the conclusion of the Mean Value Theorem.

$f\left(x\right)=\mathrm{ln}({x}^{2}+1),[a,b]=[0,1]$

The function satisfies the Mean value theorem and the value of c is 0.4028.

$f\left(x\right)=\mathrm{ln}({x}^{2}+1),[a,b]=[0,1]$

The function $f\left(x\right)=\mathrm{ln}({x}^{2}+1)$ is continuous and differentiable on $[0,1]$. The Mean Value Theorem applies to this function on the interval $[0,1]$.

The slope of the line from (0, f(0)) to (1, f(1)) is:

$\begin{array}{r}{f}^{\mathrm{\prime}}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}\\ =\frac{\mathrm{ln}\left({1}^{2}+1\right)-\mathrm{ln}\left({0}^{2}+1\right)}{1}\\ =\mathrm{ln}2-\mathrm{ln}0\\ =0.6931\end{array}$

Now,

$\begin{array}{r}{f}^{\mathrm{\prime}}\left(x\right)=\frac{d}{dx}\mathrm{ln}\left({x}^{2}+1\right)\\ =\frac{1}{{x}^{2}+1}\cdot 2x\\ =\frac{2x}{{x}^{2}+1}\end{array}$

Therefore,

$\begin{array}{r}{f}^{\mathrm{\prime}}\left(c\right)=0.6931\\ \frac{2c}{{c}^{2}+1}=0.6931\\ 2c=0.6931{c}^{2}+0.6931\end{array}$

$\begin{array}{r}0.6931{c}^{2}-2c+0.6931=0\\ {c}^{2}-2.8855c+1=0\end{array}$

role="math" localid="1648380951546" $\begin{array}{r}c=\frac{-(-2.8855)\pm \sqrt{(-2.8855{)}^{2}-4\times 1\times 1}}{2\cdot 1}\\ c=\frac{2.8855\pm \sqrt{4.3261}}{2}\\ =\frac{2.8855\pm 2.08}{2}\\ c=2.48275or0.40275\end{array}$

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