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

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

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extreme f(x)=(x^2)/(x^2+3),-1<= x<= 1

extreme points

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

Maximum (- 1, \frac{1}{4}), Minimum (0, 0), Maximum (1, \frac{1}{4})

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1

Domain of

\frac{x ^{2}}{x ^{2} + 3}, - 1 \leq x \leq 1 : - 1 \leq x \leq 1

Find intervals:

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1

Plug the extreme point

x = - 1

into

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

Maximum (- 1, \frac{1}{4})

Plug the extreme point

x = 0

into

\frac{x ^{2}}{x ^{2} + 3} \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

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

Maximum (1, \frac{1}{4})

Maximum (- 1, \frac{1}{4}), Minimum (0, 0), Maximum (1, \frac{1}{4})

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

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

f (x) = x 1 - 5 x

f (x, y) = 9 x^{2} + 3 x y + 9 y^{2} + 7 x y^{2} + 3 x^{2} y

y = - 5 x^{2} - 6 x + 2

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