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Let $f$ be a function such that: $f: [\pi/4,\,\pi/2]\to \mathbb{R}$ $$ f(x)=\begin{cases} \frac{1}{\tan{x}}& \pi/4\leq x<\pi/2, \\ 0 & x=\pi/2. \end{cases} $$ How can I show that $f$ is differentiable on the closed interval $[\pi/4,\,\pi/2]$? I can easily show continuity but I’m not sure what to do to show it’s differentiable on the closed interval.

Thank you!

Enforce
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No Name
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  • Do you know one side differentiability? https://en.wikipedia.org/wiki/Semi-differentiability – zkutch Nov 11 '24 at 00:36
  • The typical approach here would be to calculate the derivative away from your "problem point" (in this case $\pi/2$), then calculate the derivative explicitly via the definition at $\pi/2$ and show that the resulting function is continuous. Have you tried that? – Enforce Nov 11 '24 at 00:36
  • @zkutch I don't think it matters in this case because the function is only defined on the closed interval, so the derivaitve at $\pi/2$ is the one-sided derivative. – Enforce Nov 11 '24 at 00:39

2 Answers2

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$f\left(\frac{\pi}{2}\right)=0$ is used significantly in the following formula: $$\frac{f(x)-f\left(\frac{\pi}{2}\right)}{x-\frac{\pi}{2}}=\frac{\frac{1}{\tan x}-0}{x-\frac{\pi}{2}} = -1\cdot \frac{\sin\left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right) }\cdot\frac{1}{\sin x}$$

zkutch
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  • Do you know $\frac{\sin y}{y}\to 1, y\to 0$? – zkutch Nov 11 '24 at 02:01
  • Yes, but I don’t understand how it’s helpful since I already know the left side derivative at $\pi/2,$ which is equal to $-1.$ The only issue I have is with concluding that the function is differentiable over the closed interval. – No Name Nov 11 '24 at 02:03
  • Consider $x\to\frac{\pi}{2}, x< \frac{\pi}{2}$, then it is definition of derivative from left. – zkutch Nov 11 '24 at 02:56
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$ \cot x $ is a continuous ( not piecewise) function including interval $(x,0, \pi)$.

By differentiation its derivative $ -\csc^2(x)=-1$ at point P $ (\pi/2,0) $ which is a point anyhow right the curve without any slope change.

Narasimham
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