Calculating absolute min/max

[lea.g]

Wondering if someone can possibly help me out with this question. Find the absolute maximum and absolute minimum values of f(x) on the given interval.
f(x) = 3x^(1/3)-x for the interval [-2,2]. I've calculated the derivate f'(x) = (1/x^(2/3))-1 and found the critical point of x=1. I also know that my absolute maximum is at (1,2), but when I graph the equation is looks like I should have an absolute minimum at x=-1, but I'm not sure how to calculate it. I'm not seeing how to get it as a critical point.

Thanks!

 

[lookagain]

Find the absolute maximum and absolute minimum values of f(x) on the given interval.

f(x) = 3x^(1/3) - x for the interval [-2,2]. I've calculated the derivate f'(x) = (1/x^(2/3)) - 1
and found the critical point of x=1. I also know that my absolute maximum is at (1,2),
but when I graph the equation is looks like I should have an absolute minimum at x=-1,
but I'm not sure how to calculate it. I'm not seeing how to get it as a critical point.

\(\displaystyle \dfrac{1}{x^{2/3}} \ - \ 1 \ = \ 0\)


\(\displaystyle \dfrac{1}{x^{2/3}} \ = \ 1 \ \)


\(\displaystyle 1 \ = \ x^{2/3} \)


\(\displaystyle (1)^3 \ = \ \bigg(x^{2/3}\bigg)^3 \)


\(\displaystyle 1 \ = x^2\)


\(\displaystyle \pm\sqrt{1} \ = \ x\)


\(\displaystyle \pm1 \ = \ x \)


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x = 0 is also a critical number. The derivative is undefined there. The critical point there is (0, 0).

 

[lea.g]

Thanks so much! I'm not sure how I missed the that when taking the square root!