What is the extreme (x+6)/(24-sqrt(x^2-49)) ?

The extreme (x+6)/(24-sqrt(x^2-49)) is Maximum(-7,-1/24),Minimum(7, 13/24)

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extreme (x+6)/(24-sqrt(x^2-49))

extreme points

\frac{x + 6}{24 - x ^{2} - 49}

Maximum (- 7, - \frac{1}{24}), Minimum (7, \frac{13}{24})

Solution steps

f^{'} (x) = \frac{6 x + 24 x ^{2} - 49 + 49}{( 24 - x ^{2} - 49 ) ^{2} x ^{2} - 49}

Find intervals:

Increasing

: - \infty < x < - 25,

Increasing

: - 25 < x < - 7,

Increasing

: 7 < x < 25,

Increasing

: 25 < x < \infty

Plug

x = - 7

into

\frac{x + 6}{24 - x ^{2} - 49} : - \frac{1}{24}

Maximum (- 7, - \frac{1}{24})

Plug

x = 7

into

\frac{x + 6}{24 - x ^{2} - 49} : \frac{13}{24}

Minimum (7, \frac{13}{24})

Maximum (- 7, - \frac{1}{24}), Minimum (7, \frac{13}{24})

x^{2} + y^{2} + x^{- 2} y^{- 2} + 9

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

f (x) = x^{2} - 9, - 3 \leq x \leq 4

y = \frac{x}{x ^{2} - x + 1}

f (x) = 4 x^{3} - 12 x^{2} + 8 x

What is the extreme (x+6)/(24-sqrt(x^2-49)) ?
The extreme (x+6)/(24-sqrt(x^2-49)) is Maximum(-7,-1/24),Minimum(7, 13/24)