You should notice the following features of these solutions:

• In \(g(x)=2f(x)\text{,}\) the “2” simply doubled all of the \(y\)-values from \(f(x)\text{.}\)

• In \(j(x)=f(2x)\text{,}\) the “2” actually cut all of the \(x\)-values in half from \(f(x)\text{.}\) This is potentially contrary to what you might expect. Verify this by substituting values in for \(x\text{.}\)

• In \(h(x)=f(x)+1\text{,}\) the “\(+1\)” simply adds 1 unit to all of the \(y\)-values from \(f(x)\text{.}\)

• In \(k(x)=f(x-1)\text{,}\) the “\(-1\)” actually moves the graph of \(f(x)\) to the right. This is potentially contrary to what you might expect. Verify this by substituting values in for \(x\text{.}\)

1. The function \(g(x)\) should change the sign on all of the \(y\) values of \(f\). The function \(h(x)\) will shift all of the \(y\) values of \(f\) down 1 unit.

2. The order in which you do the transformations does matter. In this problem the order of operations should be to change the sign on the \(y\) value and then to shift 1 unit down.

To evaluate \(f(g(3))\) we consider that \(g(3) = 2\) and \(g(2) = 4\text{.}\) Therefore, \(f(g(3))=4\text{.}\) Similarly, \(g(f(3)) = g(9) = 8\text{.}\) The function compositions \(f(g(x))\) and \(g(f(x))\) are \(f(g(x)) = (x-1)^2\) and \(g(f(x))=x^2 - 1\text{.}\) Notice the difference between these resulting functions; the order that the composition takes places matters!

1. For \(f(x) = x^2\) and \(g(x) = x+8\) \begin{equation*} f(g(3)) = f(11) = 121, \quad g(f(3)) = g(9) = 17, \end{equation*} \begin{equation*} f(g(x)) = f(x+8) = (x+8)^2, \quad g(f(x)) = g(x^2) = x^2 + 8, \end{equation*} \begin{equation*} f(x)g(x) =(x^2)(x+8) = x^3 + 8x^2 \end{equation*}

2. From the table we have

   \begin{equation*} f(-3) = 3, \quad g(3) = 2, \quad f(g(-3)) = f(-2) = 1, \end{equation*} \begin{equation*} f(g(f(-3))) = f(g(3)) = f(2) = 1 \end{equation*}

3. \begin{equation*} f(1) = 1, \quad g(2) = 1, \quad g(f(1)) = g(1) = -1 \end{equation*} \begin{equation*} f(g(1)) = f(-1) = -2, \quad g(f(f(0))) = g(f(-2)) = g(1) = 1 \end{equation*}

1. \(f(x)=x^2\) is an even function. If you reflect of the \(y\) axis you get a perfect match.

2. \(g(x) = x^3\) is an odd function. If you rotate about the origin you get a perfect match.

3. If \(f(x)\) is even then \(f(-x) = f(x)\). If \(f(x)\) is odd then \(f(-x) = -f(x)\).

4. Here we will use the test from part (3).

   \begin{equation*} h(-x) = \frac{1}{(-x)} = -\frac{1}{x} = -h(x) \implies h(x) \text{ is odd } \end{equation*} \begin{equation*} j(-x) = e^{-x} = \frac{1}{e^x} \implies j(x) \text{ is neither even nor odd } \end{equation*} \begin{equation*} k(-x) = (-x)^2 - (-x)^4 = x^2 - x^4 = k(x) \implies k(x) \text{ is even } \end{equation*} \begin{equation*} n(-x) = (-x)^3 + (-x)^2 = -x^3 + x^2 \implies n(x) \text{ is neither even nor odd } \end{equation*}

5. See the plot below for the answers to the left and middle plots. The right-hand plot can be anything other than the first two.

   

1. To find the inverse of \(f(x)\) we first interchange the \(x\) and \(y\text{.}\) Then we solve for \(y\text{.}\) That is:

   \begin{equation*} \text{ Solve for \(y\): } x = y^2+1 \implies y = \pm\sqrt{x-1}. \end{equation*}

   Obviously the resulting solution is two equations. By convention we choose the positive square root and note that the inverse only makes sense if \(x\ge 1\text{.}\) Hence, in order for the inverse to make sense we need a restriction on the domain of \(f(x)\text{:}\) If \(0 \le x \lt \infty\) then any horizontal line only crosses the graph of \(f(x)\) once, and hence the inverse exists and is unique.

   \begin{equation*} f^{-1}(x) = \sqrt{x-1}, x \ge 1 \end{equation*}

2. Interchanging the \(x\) and \(y\) in this equation gives

   \begin{equation*} \text{ Solve for \(y\): } x = ay+b \implies y = \frac{x-b}{a}. \end{equation*}

   There is no need to restrict the domain of \(g(x)\) in this instance since the resulting equation is a function.

   \begin{equation*} g^{-1}(x) = \frac{x-b}{a} = \frac{1}{a} x - \frac{b}{a} \end{equation*}

3. Interchanging the \(x\) and \(y\) in this equation gives

   \begin{equation*} \text{ Solve for \(y\): } x = (2y+8)^3 \implies y = \frac{x^{1/3}-8}{2}. \end{equation*}

   In this instance there is no restriction on the domain of \(h(x)\) since (as in part (b)) the resulting equation is a function.

   \begin{equation*} h^{-1}(x) = \frac{x^{1/3}-8}{2} \end{equation*}

1. • For \(y=\sqrt{x-1}\): First switch the \(x\) and \(y\) to get \(x = \sqrt{y-1}\). Then if we square both sides and add 1 we get \(y = x^2 + 1\). Notice graphically that the inverse function only makes sense for \(x \ge 0\).

   • For \(y = -\frac{1}{3}x + 1\): First switch the \(x\) and \(y\) to get \(x = -\frac{1}{3}y + 1\). Then if we subtract \(1\) and multiply by \(-3\) we get \(y = -3(x-1) = -3x+3\). This linear function makes sense for all \(x\).

   • For \(y=\frac{x+4}{2x-5}\): First switch the \(x\) and \(y\) to get \(x = \frac{y+4}{2y-5}\). Multiplying by the denominator gives \(x(2y-5) = y+4\), and then if we distribute the \(x\) we get \(2xy - 5x = y+4\). Now we should gather all of the \(y\) terms on the same side to get \(2xy - y = 5x + 4\). Factoring the \(y\) out from the left and dividing gives \(y = \frac{5x+4}{2x-1}\). This function makes sense for all \(x\) not equal to \(1/2\).

2. If we compose the two functions then \begin{equation*} f(g(x)) = 3\left( \frac{x}{3} + \frac{2}{3} \right) -2 = x + 2 -2 = x. \end{equation*} \begin{equation*} g(f(x)) = \frac{1}{3} \left( 3x-2 \right) + \frac{2}{3} = x - \frac{2}{3} + \frac{2}{3} = x. \end{equation*} Since both \(f(g(x)) = x\) and \(g(f(x)) = x\) we know that \(f\) and \(g\) are inverses of each other.