What is the extreme f(x)=x^3-5x^2+8x ?

The extreme f(x)=x^3-5x^2+8x is Maximum(4/3 , 112/27),Minimum(2,4)

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extreme f(x)=x^3-5x^2+8x

extreme points

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

Maximum (\frac{4}{3}, \frac{112}{27}), Minimum (2, 4)

Solution steps

f^{'} (x) = 3 x^{2} - 10 x + 8

Find intervals:

Increasing

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

Decreasing

: \frac{4}{3} < x < 2,

Increasing

: 2 < x < \infty

Plug

x = \frac{4}{3}

into

x^{3} - 5 x^{2} + 8 x : \frac{112}{27}

Maximum (\frac{4}{3}, \frac{112}{27})

Plug

x = 2

into

x^{3} - 5 x^{2} + 8 x : 4

Minimum (2, 4)

Maximum (\frac{4}{3}, \frac{112}{27}), Minimum (2, 4)

f (x, y) = ln (2 x + 2 y)

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

(x + 1)^{5} - 5 x - 2

f (x) = \frac{2 x - 1}{x ^{2}}

f (x, y) = (2 x - x^{2}) \cdot (2 y - y^{2})

What is the extreme f(x)=x^3-5x^2+8x ?
The extreme f(x)=x^3-5x^2+8x is Maximum(4/3 , 112/27),Minimum(2,4)