# How to find singular points of a function without knowing the graph?

Problem: Let $f(x) = (x-1)^{2/3} - (x+1)^{2/3}$. Locate and classify all local extreme values of this function. Determine whether any of these extreme values are absolute.

Attempt at solution: We have $f'(x) = \frac{2}{3} (x-1)^{-1/3} - \frac{2}{3} (x+1)^{-1/3} = \frac{2}{3} [(x-1)^{-1/3} -(x+1)^{-1/3} ]$. The critical points are found when the derivative is zero. This is when $(x-1)^{-1/3} -(x+1)^{-1/3} = 0$ or $(x-1) = (x+1)$ which is a contradiction, so there are no critical points.

Now, a function can achieve a absolute maximum or minimum at critical points, endpoints or singular points. My textbook says the singular points here are $-1$ and $1$, but I don't see why ? A singular point is a point where the derivative doesn't exist. Graphically this means the graph of the function 'changes direction suddenly', and not continuously. I can just substitute $1$ in $f'(x)$ and get $f'(1) = \frac{2}{3} [(0 - (2)^{-1/3}]$ ?

So I guess my question is why these are singular points, and how I can find them without knowing the graph of the function.

• note that one of the square roots gets undefind at $x=1$ or $x=-1$ – tired Jul 18 '15 at 16:54

A critical point is defined as case 2 or 3: you can't forget to check case 2 as well. In your case, $f'(x)$ is undefined both at $x=-1$ and $x=1$, so those are critical points, and they need to be checked to see if they are extrema.
• Why is $f'(x)$ undefined at $-1$ and $-1$ ? I don't see it. I can just plug these in and I get a numerical value? – Kamil Jul 18 '15 at 17:05
• @Kamil: You can try but you will fail. For $x=1$, $$(x-1)^{-1/3}=0^{-1/3}=\frac 1{0^{1/3}}=\frac 10$$ which is undefined. Similarly for $x=-1$ and $(x+1)^{-1/3}$. Zero to any negative power is undefined. – Rory Daulton Jul 18 '15 at 17:07
• Ahh, now I see. I thought $(1-1)^{-1/3} = (0)^{-1/3} = 0$. I didn't think the exponent mattered since we have zero between the brackets, but now I understand. Thanks! – Kamil Jul 18 '15 at 17:09