What is the extreme f(x)=(x^2+x-38)/(x^2-25) ?

The extreme f(x)=(x^2+x-38)/(x^2-25) is Minimum(1, 3/2),Maximum(25, 51/50)

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extreme f(x)=(x^2+x-38)/(x^2-25)

extreme points

f (x) = \frac{x ^{2} + x - 38}{x ^{2} - 25}

Minimum (1, \frac{3}{2}), Maximum (25, \frac{51}{50})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 26 x - 25}{( x ^{2} - 25 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 5,

Decreasing

: - 5 < x < 1,

Increasing

: 1 < x < 5,

Increasing

: 5 < x < 25,

Decreasing

: 25 < x < \infty

Plug

x = 1

into

\frac{x ^{2} + x - 38}{x ^{2} - 25} : \frac{3}{2}

Minimum (1, \frac{3}{2})

Plug

x = 25

into

\frac{x ^{2} + x - 38}{x ^{2} - 25} : \frac{51}{50}

Maximum (25, \frac{51}{50})

Minimum (1, \frac{3}{2}), Maximum (25, \frac{51}{50})

f (x) = x^{\frac{4}{3}} + 4 x^{\frac{1}{3}}

h (x) = - 3 x^{5} + 5 x^{3}

f (x) = ln (5 - 2 x^{2})

f (x) = 2 x^{3} - \frac{11}{2} x^{2} - 10 x + 2

f (x, y) = 8 y^{2} + 5 x^{2} - 10 y + 6 x - 10

What is the extreme f(x)=(x^2+x-38)/(x^2-25) ?
The extreme f(x)=(x^2+x-38)/(x^2-25) is Minimum(1, 3/2),Maximum(25, 51/50)