What is the extreme f(x)=x^2sqrt(5-x) ?

The extreme f(x)=x^2sqrt(5-x) is Minimum(0,0),Maximum(4,16),Minimum(5,0)

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extreme f(x)=x^2sqrt(5-x)

extreme points

f (x) = x^{2} 5 - x

Minimum (0, 0), Maximum (4, 16), Minimum (5, 0)

Solution steps

f^{'} (x) = - \frac{5 x ( x - 4 )}{2 5 - x}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < 5

Plug

x = 0

into

x^{2} 5 - x : 0

Minimum (0, 0)

Plug

x = 4

into

x^{2} 5 - x : 16

Maximum (4, 16)

Plug

x = 5

into

x^{2} 5 - x : 0

Minimum (5, 0)

Minimum (0, 0), Maximum (4, 16), Minimum (5, 0)

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

f (x) = x^{11} - 3 x^{10} + 2

f (x, y) = y^{2} ln (x y)

f (x) = 5 sin^{2} (x), 0 \leq x \leq π

f (x, y) = x^{3} + y^{3} - 3 x^{2} - 9 y^{2} - 9 x

What is the extreme f(x)=x^2sqrt(5-x) ?
The extreme f(x)=x^2sqrt(5-x) is Minimum(0,0),Maximum(4,16),Minimum(5,0)