What is the extreme f(x)=x^3+6x^2+5 ?

The extreme f(x)=x^3+6x^2+5 is Maximum(-4,37),Minimum(0,5)

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extreme f(x)=x^3+6x^2+5

extreme points

f (x) = x^{3} + 6 x^{2} + 5

Maximum (- 4, 37), Minimum (0, 5)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 4

into

x^{3} + 6 x^{2} + 5 : 37

Maximum (- 4, 37)

Plug

x = 0

into

x^{3} + 6 x^{2} + 5 : 5

Minimum (0, 5)

Maximum (- 4, 37), Minimum (0, 5)

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

f (x, y) = (x - x^{2}) (y^{2} - 4)

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

x^{2} + y^{2} + x^{- 2} y^{- 2} + 1

f (x) = 540 x - 5 x^{3}

What is the extreme f(x)=x^3+6x^2+5 ?
The extreme f(x)=x^3+6x^2+5 is Maximum(-4,37),Minimum(0,5)