is $\arcsin(3x-4x^3)$ Non Differentiable at $x=\pm1$

Question

Find number of points at which $f(x)=\arcsin(3x-4x^3)$ is Non Differentiable in $\left[-1\:\: 1\right]$

I have just differentiated the given function and got

$$f'(x)=\frac{3(1-4x^2)}{\sqrt{1-x^2}(|1-4x^2|)}$$

So obviously $f(x)$ is Non differentiable at $x=\pm \frac{1}{2}$

But also at $x=\pm1$ we have vertical tangent, so can we also include $x=\pm1$ as Non differentiable points?

Answer 1

No, it's not obvious that $f$ is not differentiable at $\pm\frac12$ and at $\pm1$. However, it follows from what you did that$$\lim_{x\to\pm1}\bigl|f'(x)\bigr|=+\infty.$$And now, it follows from Darboux's theorem that $f$ cannot be differentiable at those points. At $\pm\frac12$, there's a different problem:$$f'(x)=\begin{cases}\frac3{\sqrt{1-x^2}}&\text{ if }-\frac12<x<\frac12\\-\frac3{\sqrt{1-x^2}}&\text{ otherwise.}\end{cases}$$So, once again, $f$ being differentiabl at $\pm\frac12$ goes against Darboux's theorem.

Answer 2

Hint:

Using Principal_values,

$$\arcsin(3x-4x^3)=\begin{cases}360^\circ n+ 3\arcsin x\ \ (1) &\mbox{if }-90^\circ\le360^\circ n+ 3\arcsin x\le90^\circ\\(2n+1)180^\circ- 3\arcsin x\ \ (2)&\mbox{if }-90^\circ\le(2n+1)180^\circ- 3\arcsin x\le90^\circ \end{cases} $$

where $n$ is any integer

$-90^\circ\le360^\circ n+ 3\arcsin x\le90^\circ\iff-30^\circ-120^\circ n\le\arcsin x\le30^\circ-120^\circ n$

$-90^\circ\le(2n+1)180^\circ- 3\arcsin x\le90^\circ\iff120^\circ n+30^\circ\le \arcsin x\le120^\circ n+90^\circ$

For $n=0,$

$$-30^\circ\le\arcsin x\le30^\circ\ \ (1)\text{ and }30^\circ\le\arcsin x\le90^\circ\ \ (2)$$

For $n=-1,$

$$90^\circ\le\arcsin x\le150^\circ\ \ \text{invalid secenario and }-90^\circ\le\arcsin x\le-30^\circ\ \ (2)$$