What is the extreme f(x)=x^3-6x^2+7 ?

The extreme f(x)=x^3-6x^2+7 is Maximum(0,7),Minimum(4,-25)

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extreme f(x)=x^3-6x^2+7

extreme points

f (x) = x^{3} - 6 x^{2} + 7

Maximum (0, 7), Minimum (4, - 25)

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} + 7 : 7

Maximum (0, 7)

Plug

x = 4

into

x^{3} - 6 x^{2} + 7 : - 25

Minimum (4, - 25)

Maximum (0, 7), Minimum (4, - 25)

y = - 3 x + 1, (3, 5)

(30.8) (40.7)

4 cos (2 (x + \frac{π}{4})) - 3

2 x + 2

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

What is the extreme f(x)=x^3-6x^2+7 ?
The extreme f(x)=x^3-6x^2+7 is Maximum(0,7),Minimum(4,-25)