What is the extreme f(x)=x^3+6x^2+16 ?

The extreme f(x)=x^3+6x^2+16 is Maximum(-4,48),Minimum(0,16)

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extreme f(x)=x^3+6x^2+16

extreme points

f (x) = x^{3} + 6 x^{2} + 16

Maximum (- 4, 48), Minimum (0, 16)

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} + 16 : 48

Maximum (- 4, 48)

Plug

x = 0

into

x^{3} + 6 x^{2} + 16 : 16

Minimum (0, 16)

Maximum (- 4, 48), Minimum (0, 16)

(- 10, 3), (2, 8)

\frac{4}{x ^{2} - 1}

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

\frac{x ^{3} + 3}{x}

f (x) = \frac{2}{x + 1}

What is the extreme f(x)=x^3+6x^2+16 ?
The extreme f(x)=x^3+6x^2+16 is Maximum(-4,48),Minimum(0,16)