What is the extreme f(x)= x/(5x+4),7<= x<= 13 ?

The extreme f(x)= x/(5x+4),7<= x<= 13 is Minimum(7, 7/39),Maximum(13, 13/69)

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extreme f(x)= x/(5x+4),7<= x<= 13

extreme points

f (x) = \frac{x}{5 x + 4}, 7 \leq x \leq 13

Minimum (7, \frac{7}{39}), Maximum (13, \frac{13}{69})

Solution steps

Find the critical points:

x = 7, x = 13

Domain of

\frac{x}{5 x + 4}, 7 \leq x \leq 13 : 7 \leq x \leq 13

Find intervals:

Increasing

: 7 < x < 13

Plug the extreme point

x = 7

into

\frac{x}{5 x + 4} \Rightarrow y = \frac{7}{39}

Minimum (7, \frac{7}{39})

Plug the extreme point

x = 13

into

\frac{x}{5 x + 4} \Rightarrow y = \frac{13}{69}

Maximum (13, \frac{13}{69})

Minimum (7, \frac{7}{39}), Maximum (13, \frac{13}{69})

f (x, y) = x y^{2} + x^{2} + y + 4

f (x) = x + 3

f (x) = x + 1

f (x y) = x y + 3 x

(x + 1)^{9} - 9 x - 2

What is the extreme f(x)= x/(5x+4),7<= x<= 13 ?
The extreme f(x)= x/(5x+4),7<= x<= 13 is Minimum(7, 7/39),Maximum(13, 13/69)