What is the extreme f(x)=(x^2-4)/(x^2+4),-5<= x<= 5 ?

The extreme f(x)=(x^2-4)/(x^2+4),-5<= x<= 5 is Maximum(-5, 21/29),Minimum(0,-1),Maximum(5, 21/29)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(x^2-4)/(x^2+4),-5<= x<= 5

extreme points

f (x) = \frac{x ^{2} - 4}{x ^{2} + 4}, - 5 \leq x \leq 5

Maximum (- 5, \frac{21}{29}), Minimum (0, - 1), Maximum (5, \frac{21}{29})

Solution steps

Find the critical points:

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

Domain of

\frac{x ^{2} - 4}{x ^{2} + 4}, - 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} - 4}{x ^{2} + 4} \Rightarrow y = \frac{21}{29}

Maximum (- 5, \frac{21}{29})

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 = 5

into

\frac{x ^{2} - 4}{x ^{2} + 4} \Rightarrow y = \frac{21}{29}

Maximum (5, \frac{21}{29})

Maximum (- 5, \frac{21}{29}), Minimum (0, - 1), Maximum (5, \frac{21}{29})

f (x) = 7 x^{2} - 56 x + 84

y = - 4 x^{2} - 5 x + 1

f (x) = \frac{x ^{2} - 7 x - 6}{x - 5}

(x + 1)^{3} (x - 3)

f (x) = 2 x^{3} + 9 x^{2} - 60 x + 6, - 5 \leq x \leq 5

What is the extreme f(x)=(x^2-4)/(x^2+4),-5<= x<= 5 ?
The extreme f(x)=(x^2-4)/(x^2+4),-5<= x<= 5 is Maximum(-5, 21/29),Minimum(0,-1),Maximum(5, 21/29)