What is the extreme f(x)=(x^2-4)/(x^2+4),-4<= x<= 4 ?

The extreme f(x)=(x^2-4)/(x^2+4),-4<= x<= 4 is Maximum(-4, 3/5),Minimum(0,-1),Maximum(4, 3/5)

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extreme f(x)=(x^2-4)/(x^2+4),-4<= x<= 4

extreme points

f (x) = \frac{x ^{2} - 4}{x ^{2} + 4}, - 4 \leq x \leq 4

Maximum (- 4, \frac{3}{5}), Minimum (0, - 1), Maximum (4, \frac{3}{5})

Solution steps

Find the critical points:

x = - 4, x = 0, x = 4

Domain of

\frac{x ^{2} - 4}{x ^{2} + 4}, - 4 \leq x \leq 4 : - 4 \leq x \leq 4

Find intervals:

Decreasing

: - 4 < x < 0,

Increasing

: 0 < x < 4

Plug the extreme point

x = - 4

into

\frac{x ^{2} - 4}{x ^{2} + 4} \Rightarrow y = \frac{3}{5}

Maximum (- 4, \frac{3}{5})

Plug the extreme point

x = 0

into

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

Minimum (0, - 1)

Plug the extreme point

x = 4

into

\frac{x ^{2} - 4}{x ^{2} + 4} \Rightarrow y = \frac{3}{5}

Maximum (4, \frac{3}{5})

Maximum (- 4, \frac{3}{5}), Minimum (0, - 1), Maximum (4, \frac{3}{5})

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

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

f (x) = x^{2} e^{2 x}

y = 2 x^{3} + 3 x^{2} - 12 x

f (x) = y - x

What is the extreme f(x)=(x^2-4)/(x^2+4),-4<= x<= 4 ?
The extreme f(x)=(x^2-4)/(x^2+4),-4<= x<= 4 is Maximum(-4, 3/5),Minimum(0,-1),Maximum(4, 3/5)