What is the extreme f(x)=x^3+6x^2+1 ?

The extreme f(x)=x^3+6x^2+1 is Maximum(-4,33),Minimum(0,1)

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extreme f(x)=x^3+6x^2+1

extreme points

f (x) = x^{3} + 6 x^{2} + 1

Maximum (- 4, 33), Minimum (0, 1)

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} + 1 : 33

Maximum (- 4, 33)

Plug

x = 0

into

x^{3} + 6 x^{2} + 1 : 1

Minimum (0, 1)

Maximum (- 4, 33), Minimum (0, 1)

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

f (x) = \frac{1}{4 - x ^{2}}

f (x) = x^{4} - 50 x^{2} + 11

(- \frac{1}{2}, \frac{3}{4}), (\frac{7}{2}, - \frac{13}{4})

sin (5 x + \frac{π}{2})

What is the extreme f(x)=x^3+6x^2+1 ?
The extreme f(x)=x^3+6x^2+1 is Maximum(-4,33),Minimum(0,1)