The Mean Value Theorem — Question 6

Question 6 - Solution

The function $f(x)=\ln(1+x^2)$ is continuous on $[-1,2]$ and differentiable on $(-1,2)$, so the Mean Value Theorem applies.

$$\frac{f(2)-f(-1)}{2-(-1)}=\frac{\ln(5/2)}3,\qquad f'(c)=\frac{2c}{1+c^2}.$$

Let $L=\ln(5/2)$. Then $Lc^2-6c+L=0$, so

$$c=\frac{3\pm\sqrt{9-L^2}}{L}.$$

The plus root exceeds $2$. The only admissible value is

$$\boxed{c=\frac{3-\sqrt{9-[\ln(5/2)]^2}}{\ln(5/2)}\approx 0.156453.}$$