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Answer 1 · 2017-03-15T13:09:41

Please see below.

$f(x) = tanx$ on $[-2pi, 2pi]$ does not satisfy the hypotheses of the Mean Value Theorem. It also does not satisfy the conclusion.

The hypotheses call for a function that is continuous on $[a,b]$ (and differentiable on $(a,b)$ ).

Tangent is undefined for odd multiples of $pi/2$ , so $tanx$ is not continuous on $[-2pi,2pi]$ .

$f(x) = tanx$ on $[-2pi, 2pi]$ does not satisfy the conclusion of the Mean Value Theorem.

$f'(x) = sec^2x$ and $(f(2pi)-f(-2pi))/(2pi-(-2pi)) = (0-0)/(4pi) = 0$ .

Since $sec^2 x >= 1$ for all $x$ , there is no $c$ anywhere for which $f'(c) = (f(2pi)-f(-2pi))/(2pi-(-2pi))$