What is the extreme f(x)=(x-7)/(x^2-25),0<= x<5 ?

The extreme f(x)=(x-7)/(x^2-25),0<= x<5 is Maximum(0, 7/25),Minimum(7-2sqrt(6),(7+2sqrt(6))/(50))

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extreme f(x)=(x-7)/(x^2-25),0<= x<5

extreme points

f (x) = \frac{x - 7}{x ^{2} - 25}, 0 \leq x < 5

Maximum (0, \frac{7}{25}), Minimum (7 - 26, \frac{7 + 2 6}{50})

Solution steps

Find the critical points:

x = 0, x = 7 - 26

Domain of

\frac{x - 7}{x ^{2} - 25}, 0 \leq x < 5 : 0 \leq x < 5

Find intervals:

Decreasing

: 0 < x < 7 - 26,

Increasing

: 7 - 26 < x < 5

Plug the extreme point

x = 0

into

\frac{x - 7}{x ^{2} - 25} \Rightarrow y = \frac{7}{25}

Maximum (0, \frac{7}{25})

Plug the extreme point

x = 7 - 26

into

\frac{x - 7}{x ^{2} - 25} \Rightarrow y = \frac{7 + 2 6}{50}

Minimum (7 - 26, \frac{7 + 2 6}{50})

Maximum (0, \frac{7}{25}), Minimum (7 - 26, \frac{7 + 2 6}{50})

f (x) = x 600 - x

f (x, y) = x (y + 1) + (x + y + 1)

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

f (x) = \frac{25}{2} x^{2} - ln (x)

f (x) = - 2 x^{3} + 33 x^{2} - 144 x + 1

What is the extreme f(x)=(x-7)/(x^2-25),0<= x<5 ?
The extreme f(x)=(x-7)/(x^2-25),0<= x<5 is Maximum(0, 7/25),Minimum(7-2sqrt(6),(7+2sqrt(6))/(50))