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Expert-verified Found in: Page 249 ### Calculus

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

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

See the step by step solution

## Step 1. Given information

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

## Step 2. Proving Mean Value Theorem.

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

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" $\begin{array}{r}c=\frac{-\left(-2.8855\right)±\sqrt{\left(-2.8855{\right)}^{2}-4×1×1}}{2\cdot 1}\\ c=\frac{2.8855±\sqrt{4.3261}}{2}\\ =\frac{2.8855±2.08}{2}\\ c=2.48275or0.40275\end{array}$ ### Want to see more solutions like these? 