Where is $f(x) = |x^2(x+1)|$ differentiable? And where are they $C^1$ and $C^2$?

Question

I'm a maths student taking a real-analysis paper and I'm currently working down my problem sheet. I've been asked the above question.

First I define a piece-wise function to describe the absolute function above.

$$f(x)= \begin{cases} x^2(x+1) & x \geq -1 \\ -(x^2)(x+1) & x < -1 \\ \end{cases} $$

$$f(x)= \begin{cases} x^3+x^2 & x \geq -1 \\ -x^3-x^2 & x < -1 \\ \end{cases} $$ My guess is that since both pieces are polynomials that $f(x)$ is smooth and thus the infinity differentiable everywhere.

I'm still wrapping my head around all this $C^1$ and $C^2$, smoothness, etc. So if anyone has any tips or tricks I'd love to know!

Thanks for your time!

Answer 1

The function $f$ is not differentiable at $-1$, since$$\lim_{x\to-1^+}\frac{f(x)-f(-1)}{x+1}=5$$and$$\lim_{x\to-1^-}\frac{f(x)-f(-1)}{x+1}=-5;$$therefore the limit $\lim_{x\to-1}\frac{f(x)-f(-1)}{x+1}$ doesn't exist. But $f$ is differentiable at every point $a\ne-1$ since:

• if $a>-1$, $\displaystyle\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{x^2(x+1)-a^2(a+1)}{x-a}=3a^2+2a$;
• if $a<-1$, $\displaystyle\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{-x^2(x+1)+a^2(a+1)}{x-a}=-3a^2-2a$;