Shifted Functions and Their Inverses | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Move the intercept and asymptote together

In $y=2+e^x$ , adding 2 moves every output up by 2. The point (0,1) becomes (0,3), and the horizontal asymptote moves from $y=0$ to $y=2$ .

For $y=\ln(x-2)$ , the log input must be positive: $x-2>0$ . Replacing $x$ by $x-2$ moves the log curve right by 2. Its vertical asymptote is $x=2$ ; its zero occurs at $x-2=1$ .

A shift moves the asymptote and every point by the same amount.

Your turnPractice

For $y=\ln(x+3)$ , state the domain, asymptote and horizontal-axis intercept.

Show worked answer

\begin{gathered}\text{domain: }x>-3\\\text{asymptote: }x=-3\\\ln(x+3)=0\\x+3=1\\\text{intercept }(-2,0)\end{gathered}

Undo the outside operations first

The notation $f^{-1}$ means inverse function, not $1/f$ . To find it, write $y=f(x)$ , solve for $x$ , then swap the variable names.

Worked example

Find the inverse of f(x) = 3eˣ + 2.

\begin{gathered}y-2=3e^x,\\\frac{y-2}{3}=e^x,\\x=\ln\left(\frac{y-2}{3}\right).\end{gathered}

The original range is $y>2$ , so the inverse domain is $x>2$ .

Answer

f^{-1}(x)=\ln\left(\frac{x-2}{3}\right)

Check the point: $f(0)=5$ and $f^{-1}(5)=\ln1=0$ .

Your turnPractice

Find the inverse of $g(x)=\ln(x-1)+2$ , where $x>1$ . State the inverse domain and range.

Show worked answer

\begin{gathered}y-2=\ln(x-1)\\x=1+e^{y-2}\end{gathered}

Answer

g^{-1}(x)=1+e^{x-2}

Domain: all real numbers. Range: outputs greater than 1. For example, $g(2)=2$ and $g^{-1}(2)=2$ .

Move the intercept and asymptote together

In $y=2+e^x$ , adding 2 moves every output up by 2. The point (0,1) becomes (0,3), and the horizontal asymptote moves from $y=0$ to $y=2$ .

For $y=\ln(x-2)$ , the log input must be positive: $x-2>0$ . Replacing $x$ by $x-2$ moves the log curve right by 2. Its vertical asymptote is $x=2$ ; its zero occurs at $x-2=1$ .

A shift moves the asymptote and every point by the same amount.

Your turnPractice

For $y=\ln(x+3)$ , state the domain, asymptote and horizontal-axis intercept.

Show worked answer

\begin{gathered}\text{domain: }x>-3\\\text{asymptote: }x=-3\\\ln(x+3)=0\\x+3=1\\\text{intercept }(-2,0)\end{gathered}

Undo the outside operations first

The notation $f^{-1}$ means inverse function, not $1/f$ . To find it, write $y=f(x)$ , solve for $x$ , then swap the variable names.

Worked example

Find the inverse of f(x) = 3eˣ + 2.

\begin{gathered}y-2=3e^x,\\\frac{y-2}{3}=e^x,\\x=\ln\left(\frac{y-2}{3}\right).\end{gathered}

The original range is $y>2$ , so the inverse domain is $x>2$ .

Answer

f^{-1}(x)=\ln\left(\frac{x-2}{3}\right)

Check the point: $f(0)=5$ and $f^{-1}(5)=\ln1=0$ .

Your turnPractice

Find the inverse of $g(x)=\ln(x-1)+2$ , where $x>1$ . State the inverse domain and range.

Show worked answer

\begin{gathered}y-2=\ln(x-1)\\x=1+e^{y-2}\end{gathered}

Answer

g^{-1}(x)=1+e^{x-2}

Domain: all real numbers. Range: outputs greater than 1. For example, $g(2)=2$ and $g^{-1}(2)=2$ .