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Popular >

extreme y=x(x+2)^2

Solution

extreme points

y = x (x + 2)^{2}

Solution

Maximum (- 2, 0), Minimum (- \frac{2}{3}, - \frac{32}{27})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

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

Increasing

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

Plug

x = - 2

into

x (x + 2)^{2} : 0

Maximum (- 2, 0)

Plug

x = - \frac{2}{3}

into

x (x + 2)^{2} : - \frac{32}{27}

Minimum (- \frac{2}{3}, - \frac{32}{27})

Maximum (- 2, 0), Minimum (- \frac{2}{3}, - \frac{32}{27})

Popular Examples

extreme-0.1t^20.8t+98.8,0<= t>= 8 extreme points

- 0.1 t^{2} 0.8 t + 98.8, 0 \leq t \geq 8

extreme f(x)=4x^2+xy+252x+4y^2-252y extreme points

f (x) = 4 x^{2} + x y + 252 x + 4 y^{2} - 252 y

extreme f(x)= t/(1+t^2)extreme points

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

extreme f(x)= 5/3 x^3-25x extreme points

f (x) = \frac{5}{3} x^{3} - 25 x

extreme f(x)=(x^2)/(x-8),-2<= x<=-1/4 extreme points

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

Frequently Asked Questions (FAQ)

What is the extreme y=x(x+2)^2 ?
The extreme y=x(x+2)^2 is Maximum(-2,0),Minimum(-2/3 ,-32/27)