What is the f(x)=sqrt((x^2-4x+5)/(x^2+2x+3)) ?

The f(x)=sqrt((x^2-4x+5)/(x^2+2x+3)) is

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f(x)=sqrt((x^2-4x+5)/(x^2+2x+3))

f (x) = \frac{x ^{2} - 4 x + 5}{x ^{2} + 2 x + 3}

Domain : - \infty < x < \infty

Range : \frac{6 - 34}{2} \leq f (x) \leq \frac{6 + 34}{2}

Y Intercepts : (0, \frac{5}{3})

Asymptotes : Horizontal y = 1

Extreme Points : Maximum \frac{1 - 34}{3}, \frac{6 + 34}{2}, Minimum \frac{1 + 34}{3}, \frac{6 - 34}{2}

Interval Notation

Domain : (- \infty, \infty)

Range : \frac{6 - 34}{2}, \frac{6 + 34}{2}

Solution steps

Domain of

\frac{x ^{2} - 4 x + 5}{x ^{2} + 2 x + 3} : - \infty < x < \infty

Range of

\frac{x ^{2} - 4 x + 5}{x ^{2} + 2 x + 3} : \frac{6 - 34}{2} \leq f (x) \leq \frac{6 + 34}{2}

Axis interception points of

\frac{x ^{2} - 4 x + 5}{x ^{2} + 2 x + 3} :

Y Intercepts

: (0, \frac{5}{3})

Asymptotes of

\frac{x ^{2} - 4 x + 5}{x ^{2} + 2 x + 3} :

Horizontal

: y = 1

Extreme Points of

\frac{x ^{2} - 4 x + 5}{x ^{2} + 2 x + 3} :

Maximum

\frac{1 - 34}{3}, \frac{6 + 34}{2},

Minimum

\frac{1 + 34}{3}, \frac{6 - 34}{2}

f (x) = \frac{2}{e ^{x}}

y = \frac{1}{x - 2} - 3 x

y = (ln (x))^{l n (x)}

y = 4 x + 22

f (m) = m^{2} - 4 m - 5

What is the f(x)=sqrt((x^2-4x+5)/(x^2+2x+3)) ?
The f(x)=sqrt((x^2-4x+5)/(x^2+2x+3)) is