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

The extreme f(x)=x^3-2x^2+4 is Maximum(0,4),Minimum(4/3 , 76/27)

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

extreme points

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

Maximum (0, 4), Minimum (\frac{4}{3}, \frac{76}{27})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{4}{3},

Increasing

: \frac{4}{3} < x < \infty

Plug

x = 0

into

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

Maximum (0, 4)

Plug

x = \frac{4}{3}

into

x^{3} - 2 x^{2} + 4 : \frac{76}{27}

Minimum (\frac{4}{3}, \frac{76}{27})

Maximum (0, 4), Minimum (\frac{4}{3}, \frac{76}{27})

f (x) = \frac{4 x}{x ^{2} + 9}, - 4 \leq x \leq 4

- 3 (x + 5)^{3}

f (x) = x^{3} - 6 x^{2} + 9 x + 8

f (x) = 2 x + cos (2 x), 0 < x < π

f (x, y) = x^{2} - 3 x y + 5 x - 2 y + 6 y^{2} + 8

What is the extreme f(x)=x^3-2x^2+4 ?
The extreme f(x)=x^3-2x^2+4 is Maximum(0,4),Minimum(4/3 , 76/27)