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

The extreme f(x)=2x^3+x^2-4x+4 is Maximum(-1,7),Minimum(2/3 , 64/27)

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

extreme points

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

Maximum (- 1, 7), Minimum (\frac{2}{3}, \frac{64}{27})

Solution steps

f^{'} (x) = 6 x^{2} + 2 x - 4

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < \frac{2}{3},

Increasing

: \frac{2}{3} < x < \infty

Plug

x = - 1

into

2 x^{3} + x^{2} - 4 x + 4 : 7

Maximum (- 1, 7)

Plug

x = \frac{2}{3}

into

2 x^{3} + x^{2} - 4 x + 4 : \frac{64}{27}

Minimum (\frac{2}{3}, \frac{64}{27})

Maximum (- 1, 7), Minimum (\frac{2}{3}, \frac{64}{27})

f (x) = (5 + x)^{5}

f (x, y) = y (e^{x} - 1)

f (x) = \frac{x}{x ^{2} + 9}, - 4 \leq x \leq 5

x^{3} + 3 x^{2} - 24 x

- x^{3} + 4 x y - 2 y^{2} + 1

What is the extreme f(x)=2x^3+x^2-4x+4 ?
The extreme f(x)=2x^3+x^2-4x+4 is Maximum(-1,7),Minimum(2/3 , 64/27)