# Where is Rolle's Theorem true?

## f(x) = 2 tan(x/2) find the point in the interval [0, 2pi] where the conclusion of Rolle's Theorem is true

Feb 2, 2018

Given $f \left(x\right) = 2 \tan \left(\frac{x}{2}\right)$, Since the graph of tangent is not differentiable at $x = \pi \mathmr{and} 2 \pi$, Rolle's Theorem does not apply.

#### Explanation:

Rolle's Theorem applies when $f \left(x\right)$ is continuous, differentiable, and when $f \left(a\right) = f \left(b\right)$, so there exists a value $c$ such that $f ' \left(c\right) = 0$

Since the graph of tangent is not differentiable at $x = \pi \mathmr{and} 2 \pi$, Rolle's Theorem does not apply.

Feb 10, 2018

For $f \left(x\right) = 2 \tan \left(\frac{x}{2}\right)$ there is no point in the interval $\left[0 , 2 \pi\right]$ where the conclusion of Rolle's Theorem is true.

#### Explanation:

The conclusion of Rolle's Theorem involves solving $f ' \left(x\right) = 0$.

But for $f \left(x\right) = 2 \tan \left(\frac{x}{2}\right)$, we have $f ' \left(x\right) = {\sec}^{2} \left(\frac{x}{2}\right)$ and we know, from trigonometry, that ${\sec}^{2} \left(t\right) > 1$ for all real $t$.
Therefore, we cannot solve $f ' \left(x\right) = 0$ for this function, $f$.