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

The extreme f(x)=(x^2)/(x+1),-2<= x<= 2 is Maximum(-2,-4),Minimum(0,0),Maximum(2, 4/3)

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

extreme points

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

Maximum (- 2, - 4), Minimum (0, 0), Maximum (2, \frac{4}{3})

Solution steps

Find the critical points:

x = - 2, x = 0, x = 2

Domain of

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

Find intervals:

Decreasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 2

Plug the extreme point

x = - 2

into

\frac{x ^{2}}{x + 1} \Rightarrow y = - 4

Maximum (- 2, - 4)

Plug the extreme point

x = 0

into

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

Minimum (0, 0)

Plug the extreme point

x = 2

into

\frac{x ^{2}}{x + 1} \Rightarrow y = \frac{4}{3}

Maximum (2, \frac{4}{3})

Maximum (- 2, - 4), Minimum (0, 0), Maximum (2, \frac{4}{3})

f (x) = \frac{- x - 8}{10 x ^{2} - 40 x + 27}

f (x, y) = 4 x^{2} + 9 y^{2} - 2 x - 18 x y^{2}

f (x) = \frac{x ^{3}}{3} - 10 x^{2} + 96 x + 100

f (x) = - x^{3} + 9 x^{2} + 120 x - 200, x \geq 5

f (x) = x^{\frac{3}{4}} - 2 x^{\frac{3}{4}}

What is the extreme f(x)=(x^2)/(x+1),-2<= x<= 2 ?
The extreme f(x)=(x^2)/(x+1),-2<= x<= 2 is Maximum(-2,-4),Minimum(0,0),Maximum(2, 4/3)