What is the extreme f(x)=(x-4)/(x^2-7) ?

The extreme f(x)=(x-4)/(x^2-7) is Minimum(1, 1/2),Maximum(7, 1/14)

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extreme f(x)=(x-4)/(x^2-7)

extreme points

f (x) = \frac{x - 4}{x ^{2} - 7}

Minimum (1, \frac{1}{2}), Maximum (7, \frac{1}{14})

Solution steps

f^{'} (x) = \frac{- x ^{2} + 8 x - 7}{( x ^{2} - 7 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 7,

Decreasing

: - 7 < x < 1,

Increasing

: 1 < x < 7,

Increasing

: 7 < x < 7,

Decreasing

: 7 < x < \infty

Plug

x = 1

into

\frac{x - 4}{x ^{2} - 7} : \frac{1}{2}

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

Plug

x = 7

into

\frac{x - 4}{x ^{2} - 7} : \frac{1}{14}

Maximum (7, \frac{1}{14})

Minimum (1, \frac{1}{2}), Maximum (7, \frac{1}{14})

f (x, y) = x^{3} - 48 x y + 64 y^{3}

f (x) = 7 x - 14 cos (x), - 2 \leq x \leq 0

y = (x - 9) \cdot 2 x

f (x) = x y^{2} - y^{2} + \frac{1}{2} x^{2} - 5 x + 4

What is the extreme f(x)=(x-4)/(x^2-7) ?
The extreme f(x)=(x-4)/(x^2-7) is Minimum(1, 1/2),Maximum(7, 1/14)