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

The extreme f(x)=x^3+6x^2+2 is Maximum(-4,34),Minimum(0,2)

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

extreme points

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

Maximum (- 4, 34), Minimum (0, 2)

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} + 2 : 34

Maximum (- 4, 34)

Plug

x = 0

into

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

Minimum (0, 2)

Maximum (- 4, 34), Minimum (0, 2)

\frac{x ^{2} - 7}{x - 3}

Y (x) = - 1 x + 7, (5, - 8)

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

f (x) = \frac{4}{3 + x}

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

What is the extreme f(x)=x^3+6x^2+2 ?
The extreme f(x)=x^3+6x^2+2 is Maximum(-4,34),Minimum(0,2)