What is the extreme f(x)=x^3-6x^2+15 ?

The extreme f(x)=x^3-6x^2+15 is Maximum(0,15),Minimum(4,-17)

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extreme f(x)=x^3-6x^2+15

extreme points

f (x) = x^{3} - 6 x^{2} + 15

Maximum (0, 15), Minimum (4, - 17)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

x^{3} - 6 x^{2} + 15 : 15

Maximum (0, 15)

Plug

x = 4

into

x^{3} - 6 x^{2} + 15 : - 17

Minimum (4, - 17)

Maximum (0, 15), Minimum (4, - 17)

f (x) = 3 sin (4 x)

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

f (x) = 2 x + 1

(3) (3.3)

f (x) = \frac{4 x ^{2} + 1}{2 x ^{2} + 5 x - 3}

What is the extreme f(x)=x^3-6x^2+15 ?
The extreme f(x)=x^3-6x^2+15 is Maximum(0,15),Minimum(4,-17)