What is the f(x)=((sqrt(8x^2+2)))/((3x-5)) ?

The f(x)=((sqrt(8x^2+2)))/((3x-5)) is

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f(x)=((sqrt(8x^2+2)))/((3x-5))

f (x) = \frac{( 8 x ^{2} + 2 )}{( 3 x - 5 )}

Domain : x < \frac{5}{3} or x > \frac{5}{3}

Range : f (x) \leq - \frac{2 2}{109} or f (x) > \frac{2 2}{3}

Y Intercepts : (0, - \frac{2}{5})

Asymptotes : Vertical x = \frac{5}{3}, Horizontal y = \frac{2 2}{3}, y = - \frac{2 2}{3}

Extreme Points : Maximum (- \frac{3}{20}, - \frac{2 2}{109})

Interval Notation

Domain : (- \infty, \frac{5}{3}) \cup (\frac{5}{3}, \infty)

Range : (- \infty, - \frac{2 2}{109}] \cup (\frac{2 2}{3}, \infty)

Solution steps

Domain of

\frac{8 x ^{2} + 2}{3 x - 5} : x < \frac{5}{3} or x > \frac{5}{3}

Range of

\frac{8 x ^{2} + 2}{3 x - 5} : f (x) \leq - \frac{2 2}{109} or f (x) > \frac{2 2}{3}

Axis interception points of

\frac{8 x ^{2} + 2}{3 x - 5} :

Y Intercepts

: (0, - \frac{2}{5})

Asymptotes of

\frac{8 x ^{2} + 2}{3 x - 5} :

Vertical

: x = \frac{5}{3},

Horizontal

: y = \frac{2 2}{3}, y = - \frac{2 2}{3}

Extreme Points of

\frac{8 x ^{2} + 2}{3 x - 5} :

Maximum

(- \frac{3}{20}, - \frac{2 2}{109})

y = x^{4} ln (x)

f (x) = \frac{( 2 ^{x} + 1 )}{( 2 ^{x} )}

y = ln (3 x^{2} - 4 x + 5)

y = \frac{3 x ^{2} - 5}{2 x + 2}

y = (2 x + 5) (3 x - 1)

What is the f(x)=((sqrt(8x^2+2)))/((3x-5)) ?
The f(x)=((sqrt(8x^2+2)))/((3x-5)) is