What is the extreme f(x)= x/(8x+3),8<= x<= 12 ?

The extreme f(x)= x/(8x+3),8<= x<= 12 is Minimum(8, 8/67),Maximum(12, 4/33)

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extreme f(x)= x/(8x+3),8<= x<= 12

extreme points

f (x) = \frac{x}{8 x + 3}, 8 \leq x \leq 12

Minimum (8, \frac{8}{67}), Maximum (12, \frac{4}{33})

Solution steps

Find the critical points:

x = 8, x = 12

Domain of

\frac{x}{8 x + 3}, 8 \leq x \leq 12 : 8 \leq x \leq 12

Find intervals:

Increasing

: 8 < x < 12

Plug the extreme point

x = 8

into

\frac{x}{8 x + 3} \Rightarrow y = \frac{8}{67}

Minimum (8, \frac{8}{67})

Plug the extreme point

x = 12

into

\frac{x}{8 x + 3} \Rightarrow y = \frac{4}{33}

Maximum (12, \frac{4}{33})

Minimum (8, \frac{8}{67}), Maximum (12, \frac{4}{33})

f (x) = (x^{2} + 4) (16 - x^{2})

f (x) = x^{4} + x^{3} - 3 x^{2} + 1

f (x) = 9

f (x, y) = (e^{x y})

f (x) = ln (x^{3} - 1)

What is the extreme f(x)= x/(8x+3),8<= x<= 12 ?
The extreme f(x)= x/(8x+3),8<= x<= 12 is Minimum(8, 8/67),Maximum(12, 4/33)