Critical points of $f(x)=|x^4 -16|+4x-2$, $x\geq -1$

Question

As the title says, I found two critical points 1 and 2 with its derivatives, but I don't know how to find the last critical point $x=-1$, since $f(x)$ is defined when $x=-1$. I would really appreciate it if someone can explain this to me. Thank you so much!

Answer 1

According to the definition of "critical point", $x$ is a critical point of a function $f$ if the derivative of $f$ at $x$ is $0$ or undefined.

Thus, for the function

$$f(x)=\mid x^4−16 \mid +4x−2, x \geq -1$$

$x =-1$ is a critical point, because the derivative of $f$ at $x = -1$ is undefined. It is undefined, because: The derivative of $f$ at some $x = x_0$ exists iff the left derivative and right derivative both exist and are equal. And here clearly for $x = -1$, we only have right derivative but not left derivative, because $f$ is undefined when $x<-1$.

Hope this can help you.