Question

Find the absolute extrema of the function (x^2+4)^1/3 on the interval [-5,4]

Answer 1

\[f(x)=\left( x ^{2} +4\right)^{\frac{ 1 }{ 3 }}\]
\[f'(x)=\frac{ 1 }{ 3 }\left( x ^{2} +4\right)^{\frac{ -2 }{ 3 }}2x\]
\[f'\left( x \right)=\frac{ 2x }{ 3\left( x ^{2} +4\right)^{\frac{ 2 }{ 3 }} }\]
\[f'\left( x \right)=0 gives 2x=0 ,or x=0\]
Now find the value of f(x) at x=0,-5,4
maximum value is the absolute maxima,minimum value is the absolute minima.
you got it.