I came up across an exercise in which $f(x)=|2\sin(x)+\tan(x)-3x|, x \in ({-π\over2},{π\over2})$, and I was asked:
Let ${-π\over2}<α<0<β<{π\over2}$.
If the Intermediate Value theorem is not valid in the $[α,β]$ interval, prove $α=-β$.
In the solutions, the exercise said that for IVT to not be valid, there must be $f(α)=f(β)$. Why is this the case though? Isn't it that if $f(α)=f(β)$, then it's just Rolle's theorem, so again $f'(ξ)=0$? Βut it CAN be implemented?. The conditions for IVT are again, $f$ to be continuous in $[α,β]$, and differentiable in $(α,β)$.
Am I mistaken?
Best Answer
The function is continuous for any described $\alpha, \beta$. This follows because $x \in (0, \pi/2)$ we have that $f$ is strictly increasing which can be shown by differentiation. We have something similar for $x \in (-\pi/2, 0)$. For $x = 0$ the function equals 0. So the function is continuous in the described open interval.
So the only way for the IVT to not be valid is if $f(\alpha) = f(\beta)$ in the interval $[\alpha, \beta]$. This is be definition of the IVT because you can't find any value between $f(\alpha) = f(\beta)$.
We can now see that $f(\alpha) = f(-\alpha)$ because $g(x) = 2 sin(x) + tan(x) - 3x$ satisfies $g(-\alpha) = -g(\alpha)$ so that $f(\alpha) = f(-\beta) = |g(-\beta)| = |-g(\beta)| = |g(\beta)| = f(\beta)$.
It follows that if $f(\alpha) = (\beta)$ then $\alpha = - \beta$ because $f(x)$ is strictly increasing above and under $x = 0$.