What is the extreme f(x)=x^{4/5}(9-4x) ?

The extreme f(x)=x^{4/5}(9-4x) is Minimum(0,0),Maximum(1,5)

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extreme f(x)=x^{4/5}(9-4x)

extreme points

f (x) = x^{\frac{4}{5}} (9 - 4 x)

Minimum (0, 0), Maximum (1, 5)

Solution steps

f^{'} (x) = - \frac{36 ( x - 1 )}{5 x ^{\frac{1}{5}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

x^{\frac{4}{5}} (9 - 4 x) : 0

Minimum (0, 0)

Plug

x = 1

into

x^{\frac{4}{5}} (9 - 4 x) : 5

Maximum (1, 5)

Minimum (0, 0), Maximum (1, 5)

f (x) = - \frac{1}{2} x^{4} + 8 x^{3} - 32 x^{2} - 5

f (x, y) = x^{3} + y^{3} - 12 x y

f (x) = 2 x^{3} - x^{2} + 16

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

x y e^{- x^{2} - y^{2}}

What is the extreme f(x)=x^{4/5}(9-4x) ?
The extreme f(x)=x^{4/5}(9-4x) is Minimum(0,0),Maximum(1,5)