Explain why tan x does not violate Bolzano’s theorem

by RoRi

September 5, 2015

Letting $f(x) = \tan x$ we observe that

$f \left( \frac{\pi}{4} \right) = \tan \frac{\pi}{4} = 1, \quad \text{and} \quad f \left( \frac{3 \pi}{4} \right) = \tan \frac{3 \pi}{4} = -1.$

However, there is no $x \in \left[ \frac{\pi}{4}, \frac{3 \pi}{4} \right]$ such that $f(x) =0$ . Why is this not a counterexample to Bolzano’s theorem?

This does not contradict Bolzano’s theorem since $\tan x$ is not continuous on the interval $\left[ \frac{\pi}{4}, \frac{3 \pi}{4} \right]$ since it is not defined at $\frac{\pi}{2}$ .

Stumbling Robot

Explain why tan x does not violate Bolzano’s theorem

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