What is the extreme f(x)= x/(8x-3),8<= x<= 13 ?

The extreme f(x)= x/(8x-3),8<= x<= 13 is Maximum(8, 8/61),Minimum(13, 13/101)

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extreme f(x)= x/(8x-3),8<= x<= 13

extreme points

f (x) = \frac{x}{8 x - 3}, 8 \leq x \leq 13

Maximum (8, \frac{8}{61}), Minimum (13, \frac{13}{101})

Solution steps

Find the critical points:

x = 8, x = 13

Domain of

\frac{x}{8 x - 3}, 8 \leq x \leq 13 : 8 \leq x \leq 13

Find intervals:

Decreasing

: 8 < x < 13

Plug the extreme point

x = 8

into

\frac{x}{8 x - 3} \Rightarrow y = \frac{8}{61}

Maximum (8, \frac{8}{61})

Plug the extreme point

x = 13

into

\frac{x}{8 x - 3} \Rightarrow y = \frac{13}{101}

Minimum (13, \frac{13}{101})

Maximum (8, \frac{8}{61}), Minimum (13, \frac{13}{101})

f (x, y) = x^{2} + \frac{5}{2}

f (x) = \frac{x}{( x - 7 ) ^{2}}

y = \frac{e ^{x}}{x ^{2} - 3 x + 2}

g (t) = t^{4} + t^{3} + t^{2} + 1

f (x, y) = (x + y)^{3} - 12 x y

What is the extreme f(x)= x/(8x-3),8<= x<= 13 ?
The extreme f(x)= x/(8x-3),8<= x<= 13 is Maximum(8, 8/61),Minimum(13, 13/101)