What is the extreme f(x)= 1/(x^2+2x+9),-2<= x<= 1 ?

The extreme f(x)= 1/(x^2+2x+9),-2<= x<= 1 is Minimum(-2, 1/9),Maximum(-1, 1/8),Minimum(1, 1/12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)= 1/(x^2+2x+9),-2<= x<= 1

extreme points

f (x) = \frac{1}{x ^{2} + 2 x + 9}, - 2 \leq x \leq 1

Minimum (- 2, \frac{1}{9}), Maximum (- 1, \frac{1}{8}), Minimum (1, \frac{1}{12})

Solution steps

Find the critical points:

x = - 2, x = - 1, x = 1

Domain of

\frac{1}{x ^{2} + 2 x + 9}, - 2 \leq x \leq 1 : - 2 \leq x \leq 1

Find intervals:

Increasing

: - 2 < x < - 1,

Decreasing

: - 1 < x < 1

Plug the extreme point

x = - 2

into

\frac{1}{x ^{2} + 2 x + 9} \Rightarrow y = \frac{1}{9}

Minimum (- 2, \frac{1}{9})

Plug the extreme point

x = - 1

into

\frac{1}{x ^{2} + 2 x + 9} \Rightarrow y = \frac{1}{8}

Maximum (- 1, \frac{1}{8})

Plug the extreme point

x = 1

into

\frac{1}{x ^{2} + 2 x + 9} \Rightarrow y = \frac{1}{12}

Minimum (1, \frac{1}{12})

Minimum (- 2, \frac{1}{9}), Maximum (- 1, \frac{1}{8}), Minimum (1, \frac{1}{12})

g (x) = x 32 - x^{2}

x^{5} - 2

f (x) = \frac{1}{2} x^{2} + 2 x - 10

\frac{1}{s ^{2} - 1}

x^{2} + 5 x - 36

What is the extreme f(x)= 1/(x^2+2x+9),-2<= x<= 1 ?
The extreme f(x)= 1/(x^2+2x+9),-2<= x<= 1 is Minimum(-2, 1/9),Maximum(-1, 1/8),Minimum(1, 1/12)