What is the extreme f(x)=x^3+6x^2-6[-8.3] ?

The extreme f(x)=x^3+6x^2-6[-8.3] is Maximum(-4,81.8),Minimum(0,49.8)

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extreme f(x)=x^3+6x^2-6[-8.3]

extreme points

f (x) = x^{3} + 6 x^{2} - 6 [- 8.3]

Maximum (- 4, 81.8), Minimum (0, 49.8)

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} - 6 [- 8.3] : 81.8

Maximum (- 4, 81.8)

Plug

x = 0

into

x^{3} + 6 x^{2} - 6 [- 8.3] : 49.8

Minimum (0, 49.8)

Maximum (- 4, 81.8), Minimum (0, 49.8)

f (x) = \frac{8 ( x ^{2} - 1 )}{3 x ^{\frac{1}{3}}}

f (x) = x 13 - x^{2}

x^{4} - 8 x^{2} - 10

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

f (r, h) = 2 π r h + 2 π r^{2}

What is the extreme f(x)=x^3+6x^2-6[-8.3] ?
The extreme f(x)=x^3+6x^2-6[-8.3] is Maximum(-4,81.8),Minimum(0,49.8)