What is the extreme f(x)=-x^3+4x^2+3 ?

The extreme f(x)=-x^3+4x^2+3 is Minimum(0,3),Maximum(8/3 , 337/27)

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extreme f(x)=-x^3+4x^2+3

extreme points

f (x) = - x^{3} + 4 x^{2} + 3

Minimum (0, 3), Maximum (\frac{8}{3}, \frac{337}{27})

Solution steps

f^{'} (x) = - 3 x^{2} + 8 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{8}{3},

Decreasing

: \frac{8}{3} < x < \infty

Plug

x = 0

into

- x^{3} + 4 x^{2} + 3 : 3

Minimum (0, 3)

Plug

x = \frac{8}{3}

into

- x^{3} + 4 x^{2} + 3 : \frac{337}{27}

Maximum (\frac{8}{3}, \frac{337}{27})

Minimum (0, 3), Maximum (\frac{8}{3}, \frac{337}{27})

What is the extreme f(x)=-x^3+4x^2+3 ?
The extreme f(x)=-x^3+4x^2+3 is Minimum(0,3),Maximum(8/3 , 337/27)