# Find the absolute extrema of the function on the closed

Find the absolute extrema of the function on the closed interval.

minimum - ?
maximum - ?
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John Koga

Now $\frac{dy}{dx}=2{x}^{\frac{1}{3}}-2$
$=\frac{2}{{x}^{\frac{1}{3}}}-2$
$=\frac{2\left[1-{x}^{\frac{1}{3}}\right]}{{x}^{\frac{1}{3}}}$
$x=0⇒f\left(0\right)=0$
$x=-1⇒f\left(-1\right)=3+2=5$
$x=1⇒f\left(1\right)=3-2=1$
Absolute minimum $=\left(0,0\right)$
Absolute maximum $=\left(-1,5\right)$
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Joseph Fair
How do you go from setting the derivative of y to a critical point of x=0. Please show those steps. I am getting 1. Thanks
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user_27qwe
I was struggling with this too, then after some toil remembered that critical numbers also include where the slope is undefined. The x=1 we kept getting is valid. Though, It is doubled up for evaluation by also being the endpoint on our closed interval. x=0 is where the solution is undefined, which only becomes apparent after combining the fraction!