Question

For the function $f\left(x\right)=\frac{1}{x}$, the value of $c$ that satisfy the mean value theorem for $f\left(x\right)$ on the interval $[1,4]$ is:

Answer 1

The function $f\left(x\right)=\frac{1}{x}$ is continuous and differentiable on the interval $[1,4]$. Therefore, we can apply the LMVT
$f\left(1\right)=1,f\left(4\right)=\frac{1}{4}$
So, using LMVT:
$f^{\prime }\left(x\right)=-\frac{1}{x^2}$
$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}\Rightarrow -\frac{1}{c^2}=\frac{f\left(4\right)-f\left(1\right)}{4-1}$
$\Rightarrow -\frac{1}{c^2}=\frac{\frac{1}{4}-1}{3}$
$\Rightarrow -\frac{1}{c^2}=\frac{-\frac{3}{4}}{3}$
$\Rightarrow c^2=4\Rightarrow c=\pm 2$
We see that only one root $c=2$ belongs to the open interval $(1,4)$.
$\therefore c=2$