What is the extreme f(x)=(x^2)/(1+x),-1/2 <= x<= 1 ?

The extreme f(x)=(x^2)/(1+x),-1/2 <= x<= 1 is Maximum(-1/2 , 1/2),Minimum(0,0),Maximum(1, 1/2)

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extreme f(x)=(x^2)/(1+x),-1/2 <= x<= 1

extreme points

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

Maximum (- \frac{1}{2}, \frac{1}{2}), Minimum (0, 0), Maximum (1, \frac{1}{2})

Solution steps

Find the critical points:

x = - \frac{1}{2}, x = 0, x = 1

Domain of

\frac{x ^{2}}{1 + x}, - \frac{1}{2} \leq x \leq 1 : - \frac{1}{2} \leq x \leq 1

Find intervals:

Decreasing

: - \frac{1}{2} < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - \frac{1}{2}

into

\frac{x ^{2}}{1 + x} \Rightarrow y = \frac{1}{2}

Maximum (- \frac{1}{2}, \frac{1}{2})

Plug the extreme point

x = 0

into

\frac{x ^{2}}{1 + x} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

\frac{x ^{2}}{1 + x} \Rightarrow y = \frac{1}{2}

Maximum (1, \frac{1}{2})

Maximum (- \frac{1}{2}, \frac{1}{2}), Minimum (0, 0), Maximum (1, \frac{1}{2})

f (x, y) = x^{3} + y^{3} - 300 x - 243 y + 2

f (x, y) = - 6 x^{2} - 7 x y - 6 y^{2} - 18 x - 58 y + 8

f (x) = \frac{x ^{2}}{144} - \frac{y ^{2}}{81}

25 - x^{2}

f (x) = - x^{3} + x^{2} - 9 x + 8

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