Question

Can the mean value theorem be applied to $f(x) = \dfrac{1}{x}\,$ in the interval $[1,1]$?

Answer 1

Hint: Here we have to check that the mean value theorem can be applied to the function $f(x) = \dfrac{1}{x}\,$ in the interval $[ - 1,1]$ . so, first we will check that the function is continuous on the given interval and also differentiable in the given interval. If the function is continuous and differentiable we can apply the mean value theorem in the given function otherwise not.

Complete step by step answer:
According to mean value theorem if a function $f(x)$ is continuous on a closed interval $[a,b]$ and differentiable in the open interval, then there is at least one point $x = c$ on this interval, such that
$f'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}$
Here, we have the function $f(x) = \dfrac{1}{x}\,$. First, we will check that the function is continuous in the interval $[ - 1,1]$. Here $a = - 1$ and $b = 1$.

Applying the mean value theorem in the function. we get,
$ \Rightarrow f(a) = f( - 1) = \dfrac{1}{{ - 1}}$
On dividing we get,
$ \Rightarrow f( - 1) = - 1$
Now, $f(b) = f(1) = \dfrac{1}{1}$
On dividing we get,
$ \Rightarrow f(1) = 1$
Hence, we cannot apply the mean value theorem as the function does not satisfy the first condition of the theorem. If we have to apply mean value then the function should satisfy all the conditions of the theorem. If one of the conditions is not satisfied we cannot apply the mean value theorem in the function.

Hence, the function $f(x) = \dfrac{1}{x}\,$ is not continuous in the interval $[ - 1,1]$ as $f(a) \ne f(b)$.

Note: This theorem is also known as first mean value theorem or Lagrange’s mean value theorem which allows us to express the increment of a function on an interval through the value of derivative at an intermediate point of a function. Note that we can only apply the mean value theorem if all the three conditions of the theorem are satisfied by the given function if one condition remains unsatisfied we cannot apply the mean value theorem in the given function so all the three conditions need to be satisfied.