What is the f(x)=(8-2sqrt(x+11))/(1/2 x^2-1/2 x-10) ?

The f(x)=(8-2sqrt(x+11))/(1/2 x^2-1/2 x-10) is

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f(x)=(8-2sqrt(x+11))/(1/2 x^2-1/2 x-10)

f (x) = \frac{8 - 2 x + 11}{\frac{1}{2} x ^{2} - \frac{1}{2} x - 10}

Domain : - 11 \leq x < - 4 or - 4 < x < 5 or x > 5

Range : f (x) < - \frac{1}{18} or - \frac{1}{18} < f (x) < 0 or f (x) \geq - \frac{2 ( - - \frac{2 ( 23 + 4 37 )}{9} + 11 + 4 ) ( 1639 - 404 37 )}{41391}

Y Intercepts : (0, - \frac{4 - 11}{5})

Asymptotes : Vertical x = - 4, Horizontal y = 0

Extreme Points : Maximum (- 11, \frac{1}{7}), Minimum - \frac{2 ( 23 + 4 37 )}{9}, - \frac{( - 2 - \frac{2 ( 23 + 4 37 )}{9} + 11 + 8 ) ( 1639 - 404 37 )}{41391}

Interval Notation

Domain : [- 11, - 4) \cup (- 4, 5) \cup (5, \infty)

Range : (- \infty, - \frac{1}{18}) \cup (- \frac{1}{18}, 0) \cup - \frac{2 ( - - \frac{2 ( 23 + 4 37 )}{9} + 11 + 4 ) ( 1639 - 404 37 )}{41391}, \infty

Solution steps

Domain of

\frac{8 - 2 x + 11}{\frac{1}{2} x ^{2} - \frac{1}{2} x - 10} : - 11 \leq x < - 4 or - 4 < x < 5 or x > 5

Range of

\frac{8 - 2 x + 11}{\frac{1}{2} x ^{2} - \frac{1}{2} x - 10} : f (x) < - \frac{1}{18} or - \frac{1}{18} < f (x) < 0 or f (x) \geq - \frac{2 ( - - \frac{2 ( 23 + 4 37 )}{9} + 11 + 4 ) ( 1639 - 404 37 )}{41391}

Axis interception points of

\frac{8 - 2 x + 11}{\frac{1}{2} x ^{2} - \frac{1}{2} x - 10} :

Y Intercepts

: (0, - \frac{4 - 11}{5})

Asymptotes of

\frac{8 - 2 x + 11}{\frac{1}{2} x ^{2} - \frac{1}{2} x - 10} :

Vertical

: x = - 4,

Horizontal

: y = 0

Extreme Points of

\frac{8 - 2 x + 11}{\frac{1}{2} x ^{2} - \frac{1}{2} x - 10} :

Maximum

(- 11, \frac{1}{7}),

Minimum

- \frac{2 ( 23 + 4 37 )}{9}, - \frac{( - 2 - \frac{2 ( 23 + 4 37 )}{9} + 11 + 8 ) ( 1639 - 404 37 )}{41391}

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

y = cos (x), 0 \leq x \leq 2 π

y = 1 + ln (- x)

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

What is the f(x)=(8-2sqrt(x+11))/(1/2 x^2-1/2 x-10) ?
The f(x)=(8-2sqrt(x+11))/(1/2 x^2-1/2 x-10) is