What is the extreme f(x)=x^3+2x^2+1 ?

The extreme f(x)=x^3+2x^2+1 is Maximum(-4/3 , 59/27),Minimum(0,1)

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extreme f(x)=x^3+2x^2+1

extreme points

f (x) = x^{3} + 2 x^{2} + 1

Maximum (- \frac{4}{3}, \frac{59}{27}), Minimum (0, 1)

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 0 < x < \infty

Plug

x = - \frac{4}{3}

into

x^{3} + 2 x^{2} + 1 : \frac{59}{27}

Maximum (- \frac{4}{3}, \frac{59}{27})

Plug

x = 0

into

x^{3} + 2 x^{2} + 1 : 1

Minimum (0, 1)

Maximum (- \frac{4}{3}, \frac{59}{27}), Minimum (0, 1)

\frac{5 x ^{2}}{x ^{2} + 16}

f (x) = 8 x - x^{4}, - 2 \leq x \leq 1

f (x, y) = 2 x^{2} + 4 x y^{2} - 2 x y - 10 x - 2 y + 2

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

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

What is the extreme f(x)=x^3+2x^2+1 ?
The extreme f(x)=x^3+2x^2+1 is Maximum(-4/3 , 59/27),Minimum(0,1)