Q. 57 - Calculus

Question

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.

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

Step-by-Step Solution

Answer

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

Step 1. Given information

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

Step 2. Proving Mean Value Theorem.

The function $f\left(x\right)=\ln (x^2+1)$ 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:

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

Now,

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

Therefore,

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

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

$\begin{array}{r}c=\frac{-(-2.8855)\pm \sqrt{(-2.8855{)}^2-4\times 1\times 1}}{2\cdot 1}\\ c=\frac{2.8855\pm \sqrt{4.3261}}{2}\\ =\frac{2.8855\pm 2.08}{2}\\ c=2.48275 or 0.40275\end{array}$