What are the critical points of # f(x)=x^(2/3)x^2 - 16#?

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Answer 1 · 2015-12-23T04:21:22

#c=0#

Critical points (#c#) occur when #f'(c)=0# or when #f'(c)# doesn't exist.

In order to find #f'(x)#, first simplify #f(x)#.

#f(x)=x^(2/3)x^2-16#

#f(x)=x^(2/3)x^(6/3)-16#

#f(x)=x^(8/3)-16#

Use the power rule: #d/dx[x^n]=nx^(n-1)#

#f'(x)=8/3x^(5/3)#

#f'(x)=0# when #x=0# and is never undefined.

Thus, #c=0#.

graph{x^(2/3)x^2-16 [-32.06, 32.9, -20.27, 12.2]}