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Popular >

extreme f(x)=\sqrt[3]{x^2+8}

Solution

extreme points

Solution

Minimum (0, 2)

Solution steps

f^{'} (x) = \frac{2 x}{3 ( x ^{2} + 8 ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \infty

Plug

x = 0

into

Minimum (0, 2)

Popular Examples

extreme f(x)=x^{1/3}(x+4),-27<= x<= 27 extreme points

f (x) = x^{\frac{1}{3}} (x + 4), - 27 \leq x \leq 27

extreme F(x,y)=x^3+3xy^2-3y^2-3x^2+11 extreme points

F (x, y) = x^{3} + 3 x y^{2} - 3 y^{2} - 3 x^{2} + 11

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

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

extreme 40000x+30000y-8x^2-15y^2-10xy extreme points

40000 x + 30000 y - 8 x^{2} - 15 y^{2} - 10 x y

extreme f(x)=(-x^2)/(100)+x extreme points

f (x) = \frac{- x ^{2}}{100} + x

Frequently Asked Questions (FAQ)

What is the extreme f(x)=\sqrt[3]{x^2+8} ?
The extreme f(x)=\sqrt[3]{x^2+8} is Minimum(0,2)