Answer to Question #259273 in Algebra for James

Question

Find the domain and range of the following rational function.Use any notation.

f(x)=3/x-1 f(x)=2x/x-4 f(x)=x+3/5x-5 f(x)=2x+x/2x f(x)=(x2+4x+3)/x²-9

Answer 1

$f(x)=\frac{3}{x-1}$

To find domain we set, Denominator=0

$x-1=0\\x=1$

When x=1 the function is undefined,

$\space f(x)=\frac{2x}{x-4}$

To find domain we set, Denominator=0

$\implies x-4=0\\\implies x=4$

When x=4 the function is undefined,

$\therefore$ domain of $f(x)={x:x \ne 4}$

To find range,

$Y=\frac{2x}{x-4}\\$

$\implies \frac{1}{Y}=\frac{x-4}{2x}\\$

$\implies \frac{1}{Y}=\frac{1}{2}-\frac{2}{x}$

$\implies \frac{2}{x}=\frac{1}{2}-\frac{1}{Y}$

$\implies \frac{n}{2}=\frac{2Y}{Y-2}$

$\implies n=\frac{4Y}{Y-2}$

When, Y=2 the function is undefined

$\therefore$ Range$=\{Y:Y\ne 2\}$

$f(x)=\frac{x+3}{5n-5}$

Denominator$=0\implies5n-5=0\\\implies n=1\\\therefore Domain=\{x|x\ne 1\}$

Now,

$Y=\frac{x+3}{5x-5}\\\implies5xY-5Y=x+3\\\implies x(5Y-1)=3+5Y\\n=\frac{3+5Y}{5Y-1}$

Set, $5Y-1=0\implies Y=\frac{1}{5}$

$\therefore$ Range of $f(x)=\{Y|Y\ne \frac{1}{5}\}$

$f(x)=\frac{2x+2}{2x}$

To find domain we set, Denominator=0

$2x=0\\x=0$

When x=0 the function is undefined,

$f(x)=\frac{2x+4x+3}{2x^2-9}$

To find domain we set, Denominator=0

$x^2-9=0\\x^2=9\\x=3$

When x=3 the function is undefined,