What is the extreme sqrt(x)+\sqrt[3]{x},0<= x<= 4 ?

The extreme sqrt(x)+\sqrt[3]{x},0<= x<= 4 is Minimum(0,0),Maximum(4,2+\sqrt[3]{4})

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extreme sqrt(x)+\sqrt[3]{x},0<= x<= 4

extreme points

Solution steps

Find the critical points:

x = 0, x = 4

Domain of

Find intervals:

Increasing

: 0 < x < 4

Plug the extreme point

x = 0

into

Minimum (0, 0)

Plug the extreme point

x = 4

into

f (x, y) = 1 + 2 x y

f (x, y) = 2 x + y + \frac{2}{x} - \frac{y ^{2}}{2}

f (x, y) = 2 x y - x^{3} - y^{2}

f (x) = 3 x^{4} + 8 x^{3} - 6 x^{2} - 24 x

x^{3} - 3 x^{2} - 9 x + 1

What is the extreme sqrt(x)+\sqrt[3]{x},0<= x<= 4 ?
The extreme sqrt(x)+\sqrt[3]{x},0<= x<= 4 is Minimum(0,0),Maximum(4,2+\sqrt[3]{4})