What is the extreme y=(x^4)/(1+x^3) ?

The extreme y=(x^4)/(1+x^3) is Maximum(-\sqrt[3]{4},-(4\sqrt[3]{4})/3),Minimum(0,0)

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extreme y=(x^4)/(1+x^3)

extreme points

y = \frac{x ^{4}}{1 + x ^{3}}

Solution steps

f^{'} (x) = \frac{4 x ^{3} + x ^{6}}{( 1 + x ^{3} ) ^{2}}

Find intervals:

Increasing

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

into

Plug

x = 0

into

\frac{x ^{4}}{1 + x ^{3}} : 0

Minimum (0, 0)

\frac{200}{x + 4 + 2 x}

f (x) = - 0.3 t^{2} + 2.4 t + 98.2

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

f (x) = 4.3 x - 0.01 x^{2} - 8

f (x) = x^{5} - 3125 x, - 7 \leq x \leq 7

What is the extreme y=(x^4)/(1+x^3) ?
The extreme y=(x^4)/(1+x^3) is Maximum(-\sqrt[3]{4},-(4\sqrt[3]{4})/3),Minimum(0,0)