What is the extreme f(x)=(2+x)/(7-x),-7<= x<= 0 ?

The extreme f(x)=(2+x)/(7-x),-7<= x<= 0 is Minimum(-7,-5/14),Maximum(0, 2/7)

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extreme f(x)=(2+x)/(7-x),-7<= x<= 0

extreme points

f (x) = \frac{2 + x}{7 - x}, - 7 \leq x \leq 0

Minimum (- 7, - \frac{5}{14}), Maximum (0, \frac{2}{7})

Solution steps

Find the critical points:

x = - 7, x = 0

Domain of

\frac{2 + x}{7 - x}, - 7 \leq x \leq 0 : - 7 \leq x \leq 0

Find intervals:

Increasing

: - 7 < x < 0

Plug the extreme point

x = - 7

into

\frac{2 + x}{7 - x} \Rightarrow y = - \frac{5}{14}

Minimum (- 7, - \frac{5}{14})

Plug the extreme point

x = 0

into

\frac{2 + x}{7 - x} \Rightarrow y = \frac{2}{7}

Maximum (0, \frac{2}{7})

Minimum (- 7, - \frac{5}{14}), Maximum (0, \frac{2}{7})

y = 3 x^{4} + 16 x^{3} = x^{3} (3 x + 16)

f (x) = - 4 x^{3} - 6 x^{2} + 72 x + 6

y = 12 x^{\frac{5}{7}} + 5 x^{\frac{12}{7}}

y = x^{3} - 3 x^{2} + 7 x - 10

f (x) = - \frac{10}{3} x^{3} - \frac{31}{2} x^{2} - 3 x + 4

What is the extreme f(x)=(2+x)/(7-x),-7<= x<= 0 ?
The extreme f(x)=(2+x)/(7-x),-7<= x<= 0 is Minimum(-7,-5/14),Maximum(0, 2/7)