What is the extreme f(x)= x/(x+3),-1<= x<= 6 ?

The extreme f(x)= x/(x+3),-1<= x<= 6 is Minimum(-1,-1/2),Maximum(6, 2/3)

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extreme f(x)= x/(x+3),-1<= x<= 6

extreme points

f (x) = \frac{x}{x + 3}, - 1 \leq x \leq 6

Minimum (- 1, - \frac{1}{2}), Maximum (6, \frac{2}{3})

Solution steps

Find the critical points:

x = - 1, x = 6

Domain of

\frac{x}{x + 3}, - 1 \leq x \leq 6 : - 1 \leq x \leq 6

Find intervals:

Increasing

: - 1 < x < 6

Plug the extreme point

x = - 1

into

\frac{x}{x + 3} \Rightarrow y = - \frac{1}{2}

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

Plug the extreme point

x = 6

into

\frac{x}{x + 3} \Rightarrow y = \frac{2}{3}

Maximum (6, \frac{2}{3})

Minimum (- 1, - \frac{1}{2}), Maximum (6, \frac{2}{3})

f (x) = 4 x y^{2} + 10 x - 4 y^{4} + 9

f (x) = x^{3} - 6 x^{2} + 9 x + 10

4 x + 1

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

f (x) = \frac{x ^{6}}{6} - ln (x)

What is the extreme f(x)= x/(x+3),-1<= x<= 6 ?
The extreme f(x)= x/(x+3),-1<= x<= 6 is Minimum(-1,-1/2),Maximum(6, 2/3)