Question

A function f (x) = 1 + $\frac{1}{x}$ is defined on the closed interval [1, 3]. A point in the interval, where the function satisfies the mean value theorem, is ______________.

Answer 1

The function $f\left(x\right)=1+\frac{1}{x}$ is defined on the interval [1, 3].

f(x) is continuous on [1, 3] and differentiable on (1, 3).

So, by mean value theorem there must exist at least one real number c ∈ (1, 3) such that

$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$

$\Rightarrow -\frac{1}{c^2}=\frac{{\displaystyle \frac{4}{3}}-2}{3-1}\left[f\left(x\right)=1+\frac{1}{x}\Rightarrow f\text{'}\left(x\right)=-\frac{1}{x^2}\right]$

$\Rightarrow -\frac{1}{c^2}=\frac{{\displaystyle -\frac{2}{3}}}{2}$

$\Rightarrow \frac{1}{c^2}=\frac{1}{3}$

$\Rightarrow c^2=3$

$\Rightarrow c=\pm \sqrt{3}$

Thus, $c=\sqrt{3}\in \left(1,3\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Hence, a point in the interval where the given function satisfies the mean value theorem is $\sqrt{3}$.


A function f (x) = 1 + $\frac{1}{x}$ is defined on the closed interval [1, 3]. A point in the interval, where the function satisfies the mean value theorem, is $\underline{\sqrt{3}}$.