Question #259273

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

1
Expert's answer
2021-11-03T17:43:32-0400

f(x)=3x1f(x)=\frac{3}{x-1}

To find domain we set, Denominator=0

x1=0x=1x-1=0\\x=1

When x=1 the function is undefined,


 f(x)=2xx4\space f(x)=\frac{2x}{x-4}

To find domain we set, Denominator=0

    x4=0    x=4\implies x-4=0\\\implies x=4

When x=4 the function is undefined,

\therefore domain of f(x)=x:x4f(x)={x:x \ne 4}

To find range,

Y=2xx4Y=\frac{2x}{x-4}\\

    1Y=x42x\implies \frac{1}{Y}=\frac{x-4}{2x}\\

    1Y=122x\implies \frac{1}{Y}=\frac{1}{2}-\frac{2}{x}

    2x=121Y\implies \frac{2}{x}=\frac{1}{2}-\frac{1}{Y}

    n2=2YY2\implies \frac{n}{2}=\frac{2Y}{Y-2}

    n=4YY2\implies n=\frac{4Y}{Y-2}

When, Y=2 the function is undefined

\therefore Range={Y:Y2}=\{Y:Y\ne 2\}


f(x)=x+35n5f(x)=\frac{x+3}{5n-5}

Denominator=0    5n5=0    n=1Domain={xx1}=0\implies5n-5=0\\\implies n=1\\\therefore Domain=\{x|x\ne 1\}

Now,

Y=x+35x5    5xY5Y=x+3    x(5Y1)=3+5Yn=3+5Y5Y1Y=\frac{x+3}{5x-5}\\\implies5xY-5Y=x+3\\\implies x(5Y-1)=3+5Y\\n=\frac{3+5Y}{5Y-1}

Set, 5Y1=0    Y=155Y-1=0\implies Y=\frac{1}{5}

\therefore Range of f(x)={YY15}f(x)=\{Y|Y\ne \frac{1}{5}\}


f(x)=2x+22xf(x)=\frac{2x+2}{2x}

To find domain we set, Denominator=0

2x=0x=02x=0\\x=0

When x=0 the function is undefined,


f(x)=2x+4x+32x29f(x)=\frac{2x+4x+3}{2x^2-9}

To find domain we set, Denominator=0

x29=0x2=9x=3x^2-9=0\\x^2=9\\x=3

When x=3 the function is undefined,


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