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

The extreme f(x)=x^3+2x^2-7x is Maximum(-7/3 , 392/27),Minimum(1,-4)

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

extreme points

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

Maximum (- \frac{7}{3}, \frac{392}{27}), Minimum (1, - 4)

Solution steps

f^{'} (x) = 3 x^{2} + 4 x - 7

Find intervals:

Increasing

: - \infty < x < - \frac{7}{3},

Decreasing

: - \frac{7}{3} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - \frac{7}{3}

into

x^{3} + 2 x^{2} - 7 x : \frac{392}{27}

Maximum (- \frac{7}{3}, \frac{392}{27})

Plug

x = 1

into

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

Minimum (1, - 4)

Maximum (- \frac{7}{3}, \frac{392}{27}), Minimum (1, - 4)

f (x, y) = x e^{- 3 x^{2} - 3 y^{2}}

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

\frac{1}{x - 2}

f (x) = - \frac{5 e ^{- x}}{5 - 3 x}

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

What is the extreme f(x)=x^3+2x^2-7x ?
The extreme f(x)=x^3+2x^2-7x is Maximum(-7/3 , 392/27),Minimum(1,-4)