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.
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$\begingroup$There are three possibilities for an extremum of a function on an interval:
- An endpoint of the interval.
- The derivative is undefined.
- The derivative is zero.
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.
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