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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = \frac{5 x ^{2} + 20 x}{2 x + 5}

Find intervals:

Increasing

: - 5 < x < - 4,

Decreasing

: - 4 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 5

into

x^{2} x + 5 : 0

Minimum (- 5, 0)

Plug

x = - 4

into

x^{2} x + 5 : 16

Maximum (- 4, 16)

Plug

x = 0

into

x^{2} x + 5 : 0

Minimum (0, 0)

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

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

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

f (x) = e^{x^{2} - 5 x - 1}

f (x) = 2 x^{3} - 3 x^{2} - 12 x + 3

f (x) = (x^{2} - 1) (x - 3)

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