What is the extreme f(x)=(3+x)/(7-x),-7<= x<= 0 ?

The extreme f(x)=(3+x)/(7-x),-7<= x<= 0 is Minimum(-7,-2/7),Maximum(0, 3/7)

extreme f(x)=(3+x)/(7-x),-7<= x<= 0

extreme points

f (x) = \frac{3 + x}{7 - x}, - 7 \leq x \leq 0

Minimum (- 7, - \frac{2}{7}), Maximum (0, \frac{3}{7})

Solution steps

Find the critical points:

x = - 7, x = 0

Domain of

\frac{3 + 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{3 + x}{7 - x} \Rightarrow y = - \frac{2}{7}

Minimum (- 7, - \frac{2}{7})

Plug the extreme point

x = 0

into

\frac{3 + x}{7 - x} \Rightarrow y = \frac{3}{7}

Maximum (0, \frac{3}{7})

Minimum (- 7, - \frac{2}{7}), Maximum (0, \frac{3}{7})

f (x) = e^{6 x^{2} + 4 y^{2} + 10}

f (x, y) = (x^{2} - y^{2}) e^{- (x^{2} + y^{2})}

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

f (x) = - 45 x^{2} + 400 x

f (x) = 3 x^{2} - 2 x + 5

What is the extreme f(x)=(3+x)/(7-x),-7<= x<= 0 ?
The extreme f(x)=(3+x)/(7-x),-7<= x<= 0 is Minimum(-7,-2/7),Maximum(0, 3/7)