What is the extreme f(x)=(x^2-1)/(x^2+1),-5<= x<= 5 ?

The extreme f(x)=(x^2-1)/(x^2+1),-5<= x<= 5 is Maximum(-5, 12/13),Minimum(0,-1),Maximum(5, 12/13)

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extreme f(x)=(x^2-1)/(x^2+1),-5<= x<= 5

extreme points

f (x) = \frac{x ^{2} - 1}{x ^{2} + 1}, - 5 \leq x \leq 5

Maximum (- 5, \frac{12}{13}), Minimum (0, - 1), Maximum (5, \frac{12}{13})

Solution steps

Find the critical points:

x = - 5, x = 0, x = 5

Domain of

\frac{x ^{2} - 1}{x ^{2} + 1}, - 5 \leq x \leq 5 : - 5 \leq x \leq 5

Find intervals:

Decreasing

: - 5 < x < 0,

Increasing

: 0 < x < 5

Plug the extreme point

x = - 5

into

\frac{x ^{2} - 1}{x ^{2} + 1} \Rightarrow y = \frac{12}{13}

Maximum (- 5, \frac{12}{13})

Plug the extreme point

x = 0

into

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

Minimum (0, - 1)

Plug the extreme point

x = 5

into

\frac{x ^{2} - 1}{x ^{2} + 1} \Rightarrow y = \frac{12}{13}

Maximum (5, \frac{12}{13})

Maximum (- 5, \frac{12}{13}), Minimum (0, - 1), Maximum (5, \frac{12}{13})

f (x, y) = \frac{1}{2} x^{4} - 3 x y + 3 y^{4}

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

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

f (x) = \frac{2}{3} x^{3} - x^{2} - 12 x + 3

y = x^{4} + 4 x^{3} - 2

What is the extreme f(x)=(x^2-1)/(x^2+1),-5<= x<= 5 ?
The extreme f(x)=(x^2-1)/(x^2+1),-5<= x<= 5 is Maximum(-5, 12/13),Minimum(0,-1),Maximum(5, 12/13)