Question

Let \(f(x) = \tan x\). Show that \(\;f(0) = f(\pi )\) but there is no number c in such that \({f^\prime }(c) = 0\) . Why does this not contradict Rolle’s Theorem?

Short answer

The answer is stated below.

Step-by-step solution

Given function

The function is \(f(x) = \tan x\).

Concept of Rolle’s Theorem

Rolle's Theorem:

“Let\(f\)be a function that satisfies the following hypothesis:

1)\(f\)is continuous on the closed interval\(\left( {a,b} \right)\).

2)\(f\)is differentiable on the open interval\((a,b)\).

3)\(f(a) = f(b)\).

Then, there is a number\(c\)in\((a,b)\)such that,\(f'(c) = 0\).”

Check the third condition of Rolle’s Theorem

Find the values of\(f(x)\)for the end points of the given interval.

Substitute,\(x = 0 \to f(x)\).

\(\begin{aligned}{c}f(0) &= \tan 0\\ &= 0\end{aligned}\)

Substitute,\(x = \pi \to f(x)\).

\(\begin{aligned}{c}\tan 0 &= 0\\f(\pi ) &= \tan \pi \;\;\\\tan (\pi ) &= 0\end{aligned}\)

Hence, it is proved that \(f(0) = f(\pi )\).

Check the second condition of Rolle’s Theorem

Find the derivative of\(f(x)\).

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}(\tan x)\;\\{(\tan x)^\prime } &= {\sec ^2}x\end{aligned}\)

Thus,\({f^\prime }(c) = {\sec ^2}c\).

Notice that\({f^\prime }(c) = 0\)has no solution.

Therefore there is no existence of a point\(c\).

The function\(f(x)\)fails to satisfy Rolle's Theorem conditions as the function\(f(x)\)is not differentiable on the open interval\((0,\pi )\)as well as not continuous on\((0,\pi )\).

Logically, if the hypotheses (conditions) are false and the conclusion is either true or false, then the statement is true.

Therefore, it does not contradict the Rolle's Theorem.