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

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

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

extreme points

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

Minimum (0, - 7), Maximum (2, - 3)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

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

Minimum (0, - 7)

Plug

x = 2

into

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

Maximum (2, - 3)

Minimum (0, - 7), Maximum (2, - 3)

\frac{x + 3}{x - 5}

f (x) = y^{6} = x^{3} - 16 x

ln (e x)

f (x) = \frac{3}{4} x + 1

\frac{( x + 2 ) ( x + 3 )}{2 ( x + 2 )}

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