Precalculus: Lecture Notes

1. The SFU ID Number Function is a one–to–one function because every member of the SFU community is associate with a different SFU ID number. (There are no two different people with the same ID number.)

2. The Birthday Date Function is not a one–to–one. There are at least two SFU students with the same birthday. (Why?)

Observe that

In other words, the function \(g\) is with the property that there are two different inputs, \(x_1=-2\) and \(x_2=2\text{,}\) with the same output, \(g(-2)=g(2)=4\text{.}\) Hence, by definition, the function \(g\) is not one–to–one.

Observe that the line \(y=4\) intersects the graph of the function \(g\) at two points, \((-2,4)\) and \((2,4)\text{.}\)

To establish if the function \(f\) is one–to–one or not one–to–one we demonstrate a more general approach.

Let \(x_1,x_2\in \mathbb{R}\) be such that \(f(x_1)=f(x_2)\text{.}\) Then

Hence, we have proved that the function \(f\) is with the property that there are no two different inputs with the same output. By definition, \(f\) is one–to–one.

Observe that any horizontal line \(y=a\) intersects the graph of the function \(f\) at exactly one point.

See Figure 1.79, Figure 1.80, and Figure 1.81.

Recall that \(f\) is one–to–one.

Since the “SFU ID Function” is a one–to–one function, its inverse, the function \(f^{-1}\text{,}\) exists and \(f^{-1}(\text{Dr. J's SFU ID Number})=\text{ Dr. J}\text{.}\)

1. The equation \(f(x)=8\) is equivalent to the equation \(3x+1=8\text{.}\) Thus \(\displaystyle x=\frac{7}{3}\text{.}\)

2. We solve the equation \(3x+1=y\) for \(x\text{.}\) (Since the number \(y\) is “given,” we treat \(x\) as the unknown.)

   From

   \begin{equation*} 3x+1=y\Leftrightarrow 3x=y-1\Leftrightarrow x=\frac{y-1}{3} \end{equation*}

   we conclude that, for the given \(y\in\mathbb{R}\text{,}\)

   \begin{equation*} f\left(\frac{y-1}{3}\right)= y. \end{equation*}

   This means that the INVERSE function is given by

   \begin{equation*} f^{-1}(y)=\frac{y-1}{3}. \end{equation*}

   Since the letter that we use to annotate the independent variable does not matter we will also write

   \begin{equation*} f^{-1}(x)=\frac{x-1}{3}. \end{equation*}

1. For \(x\geq 0\text{,}\)

   \begin{equation*} (g\circ g^{-1})(x)=g(g^{-1}(x))=g(\sqrt{x})=(\sqrt{x})^2=x. \end{equation*}

2. For \(x\geq 0\text{,}\)

   \begin{equation*} (f\circ g^{-1})(x)=f(g^{-1}(x))=f(\sqrt{x})=(\sqrt{x})^2=x. \end{equation*}

3. For \(x\geq 0\text{,}\)

   \begin{equation*} (g^{-1}\circ g)(x)=g^{-1}(g(x))=g^{-1}(x^2)=\sqrt{x^2}=x. \end{equation*}

4. For \(x\in \mathbb{R}\text{,}\)

   \begin{equation*} (g^{-1}\circ f)(x)=g^{-1}(f(x))=g^{-1}(x^2)=\sqrt{x^2}=|x|. \end{equation*}

Observe that the graph of the function \(f(x)=x\) passes the horizontal line test.

Looking at the graph of the the function \(f\) we note that it coincides with the line \(y=x\) and thus it is symmetric to itself with the respect to the line \(y=x\text{.}\) Hence \(f^{-1}(x)=x\text{.}\)

Observe that the graph of the function \(g(x)=\frac{1}{x}\text{,}\) \(x\not= 0\text{,}\) passes the horizontal line test.

Looking at the graph of the the function \(f\) we note that it coincides with the line \(y=x\) and thus it is symmetric to itself with the respect to the line \(y=x\text{.}\) Hence \(g^{-1}(x)=\frac{1}{x}\text{,}\) \(x\not= 0\text{.}\)

1. Since the function \(f\) is a rational function its domain is given by

   \begin{equation*} \{x\in \mathbb{R}:x+1\not= 0\}=\{x\in \mathbb{R}:x\not= -1\}=\mathbb{R}\backslash \{-1\}. \end{equation*}

2. Note that \(\displaystyle f(x)=\frac{x+2}{x+1}=\frac{1+(x+1)}{x+1}=\frac{1}{x+1}+1\text{.}\)

   See Figure 1.90.

   

3. Note that the graph of the function \(f\) passes the horizontal line test.

4. • Step 1: Write \(y=\frac{x+2}{x+1}.\)

   • Step 2: Interchange the roles of “\(x\)” and “\(y\)”

     \begin{equation*} x=\frac{y+2}{y+1} \end{equation*}

     and solve for \(y\text{:}\)

     \begin{equation*} x=\frac{y+2}{y+1}\Leftrightarrow x(y+1)=y+2\Leftrightarrow xy+1=y+2\Leftrightarrow xy-y=2-x \end{equation*}
     \begin{equation*} \Leftrightarrow (x-1)y=2-x\Leftrightarrow y=\frac{2-x}{x-1}. \end{equation*}

   • Step 3: The inverse function is

     \begin{equation*} f^{-1}(x)=\frac{2-x}{x-1} \end{equation*}

     with the domain \(\{x\in \mathbb{R}: x-1\not=0\}=\{x\in \mathbb{R}: x\not=1\}\text{.}\)

Observe from the graph that the function is increasing on the intervals \((-2,-1)\) and \((1,2)\) and decreasing on the interval \((-1,1)\text{.}\)