What is the extreme f(x)=1-x+(((x^3))/(3+x))^{1/2} ?

The extreme f(x)=1-x+(((x^3))/(3+x))^{1/2} is Maximum(0,1)

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

extreme points

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

Maximum (0, 1)

Solution steps

Find the critical points:

x = 0

Domain of

1 - x + (\frac{x ^{3}}{3 + x})^{\frac{1}{2}} : x < - 3 or x \geq 0

Find intervals:

Constant

: - \infty < x < - 3,

Decreasing

: 0 < x < \infty

Plug the extreme point

x = 0

into

1 - x + (\frac{x ^{3}}{3 + x})^{\frac{1}{2}} \Rightarrow y = 1

Maximum (0, 1)

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

a y = (x - 9) e^{x}

f (x, y) = x^{2} + y^{2} + 49

f (x) = x^{2} + y^{2} + 4 x - 16 y + 8

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

What is the extreme f(x)=1-x+(((x^3))/(3+x))^{1/2} ?
The extreme f(x)=1-x+(((x^3))/(3+x))^{1/2} is Maximum(0,1)