What is the extreme (3+x)/(9-x),-9<= x<= 0 ?

The extreme (3+x)/(9-x),-9<= x<= 0 is Minimum(-9,-1/3),Maximum(0, 1/3)

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extreme (3+x)/(9-x),-9<= x<= 0

extreme points

\frac{3 + x}{9 - x}, - 9 \leq x \leq 0

Minimum (- 9, - \frac{1}{3}), Maximum (0, \frac{1}{3})

Solution steps

Find the critical points:

x = - 9, x = 0

Domain of

\frac{3 + x}{9 - x}, - 9 \leq x \leq 0 : - 9 \leq x \leq 0

Find intervals:

Increasing

: - 9 < x < 0

Plug the extreme point

x = - 9

into

\frac{3 + x}{9 - x} \Rightarrow y = - \frac{1}{3}

Minimum (- 9, - \frac{1}{3})

Plug the extreme point

x = 0

into

\frac{3 + x}{9 - x} \Rightarrow y = \frac{1}{3}

Maximum (0, \frac{1}{3})

Minimum (- 9, - \frac{1}{3}), Maximum (0, \frac{1}{3})

(x - 1)^{2} (2 x - 8)

y = 4 x^{\frac{3}{4}} - x (256)

f (x) = 64 - x^{2}, - 8 \leq x \leq 8

- \frac{2 e ^{x}}{- 4 x - 1}

f (x) = \frac{4}{4 x ^{2} - 1} + b

What is the extreme (3+x)/(9-x),-9<= x<= 0 ?
The extreme (3+x)/(9-x),-9<= x<= 0 is Minimum(-9,-1/3),Maximum(0, 1/3)