MTH 103 Notes - Chapter 3 Section 7

Solution: Consider \((f\circ g)(x) = f(g(x)) = 3\left(\dfrac{3}{x}\right) = \dfrac{9}{x}\). Since this is not the same function as \(x\) we know \(f\) and \(g\) cannot be inverses of one another.

Solution: There are three parts to this question. First lets verify that \(g\) is the inverse of \(f\) $$\begin{align*} f(g(x)) &= 3\left(\dfrac{x-5}{3}\right) + 5 \\ &= (x-5)+5 = x \\ g(f(x)) & = \dfrac{(3x+5)-5}{3} \\ &= \dfrac{3x}{3} = x \end{align*}$$ Now we can sketch both of them via this desmos web app. Finally we notice that in this case the inverse of our linear function \(f(x)=3x+5\) is also a linear function. We are asked to answer the question "Will the inverse of a linear function always be a linear function?" The answer is yes and this is because of the property of inverses that says The graphs of \(f\) and \(f^{-1}\) are symmetric over the line \(y = x\). So when you reflect a line (like \(f(x)=3x+5\)) you will get another line.

Solution: Let's follow the technique: $$\begin{align*} f(x) &= \dfrac{4}{x+7} \tag{Replace \(f(x)\) with \(y\).}\\ y &= \dfrac{4}{x+7} \tag{Interchange \(x\) and \(y\).} \\ x &= \dfrac{4}{y+7} \tag{Solve for new \(y\).} \\ x(y+7) &= 4 \\ (y+7) &= \dfrac{4}{x} \\ y &= \dfrac{4}{x} - 7 \tag{Replace \(y\) with \(f^{-1}(x)\).} \\ f^{-1}(x) &= \dfrac{4}{x} - 7 \end{align*}$$ Let's verify our work by checking \(f(1) = \dfrac{4}{8} = \dfrac{1}{2}\). Also \(f^{-1}(1/2) = \dfrac{4}{1/2}-7=8-7=1\) so we have shown that \((1,1/2)\) is a point on \(f\) and \((1/2,1)\) is a point on \(f^{-1}\).

Solution: Again follow the technique: $$\begin{align*} h(x) &= 27x^3-1 \tag{Replace \(h(x)\) with \(y\).}\\ y &= 27x^3-1 \tag{Interchange \(x\) and \(y\).} \\ x &= 27y^3-1 \tag{Solve for new \(y\).} \\ x+1 &= 27y^3 \\ \dfrac{x+1}{27} &= y^3 \\ \sqrt[3]{\dfrac{x+1}{27}} &= y \\ \dfrac{\sqrt[3]{x+1}}{3} &= y \tag{Replace \(y\) with \(h^{-1}(x)\).} \\ \dfrac{\sqrt[3]{x+1}}{3} &= h^{-1}(x) \end{align*}$$

Solution: The technique! $$\begin{align*} f(x) &= \dfrac{2x+1}{3x-4} \tag{Replace \(f(x)\) with \(y\).}\\ y &= \dfrac{2x+1}{3x-4} \tag{Interchange \(x\) and \(y\).} \\ x &= \dfrac{2y+1}{3y-4} \tag{Solve for new \(y\).} \\ x(3y-4) &= 2y+1 \\ 3xy-4x &= 2y+1 \\ 3xy-2y &= 4x+1 \\ (3x-2)y &= 4x+1 \\ y &= \dfrac{4x+1}{3x-2} \tag{Replace \(y\) with \(f^{-1}(x)\).} \\ f^{-1}(x) &= \dfrac{4x+1}{3x-2} \end{align*}$$

Solution: Two parts this time. First lets find the inverse using the technique: $$\begin{align*} f(x) &= (x-3)^2 \tag{Replace \(f(x)\) with \(y\).}\\ y &= (x-3)^2 \tag{Interchange \(x\) and \(y\).} \\ x &= (y-3)^2 \tag{Solve for new \(y\).} \\ \pm\sqrt{x} &= y-3 \\ 3\pm\sqrt{x} &= y \end{align*}$$ Now remember this has to be a function so we must choose either plus or minus. Since \(x\leq 3\) we need to to choose the negative option. We can verify this with a graph too! $$\begin{align*} 3 - \sqrt{x} &= y \tag{Replace \(y\) with \(f^{-1}(x)\).} \\ 3 - \sqrt{x} &= f^{-1}(x) \end{align*}$$ Now that we have the inverse we can find its domain by observing that \(\sqrt{x}\) requires \(x\geq 0\) in order to yeild a real number output. Therefore we can conclude that the domain is \([0,\infty)\). Alternatively we could use the range of \(f\) (thanks to a property 2 of inverses) to see that the domain of \(f^{-1}\) is indeed \([0,\infty)\).

Solution: Use the technique: $$\begin{align*} C(F) &= \dfrac{5}{9}(F-32) \tag{Replace \(C(F)\) with \(y\).}\\ y &= \dfrac{5}{9}(F-32) \tag{Interchange \(F\) and \(y\).} \\ F &= \dfrac{5}{9}(y-32) \tag{Solve for new \(y\).} \\ \dfrac{9}{5} F &= y-32 \\ \dfrac{9}{5} F +32 &= y \tag{Replace \(y\) with \(C^{-1}(F)\).} \\ \dfrac{9}{5} F +32 &= C^{-1}(F) \end{align*}$$

Solution:

1. No, it does not pass HLT or the VLT
2. No, it does not pass HLT
3. Yes, it passes both HLT and the VLT
4. No, it does not pass HLT or the VLT

Solution:

1. Yes, \(f(x)\) is a line (and is not vertical or horizontal) so it is one-to-one
2. No, for instance \(f(1) = 2-7=-5\) and \(f(-1)=2-7=-5\) so there are two \(x\) values that give the same \(y\) value to \(f\) will not pass the HLT.
3. No, for instance \(h(1) =|1-3| =2\) and \(h(5)=|5-3|=2\) so there are two \(x\) values that give the same \(y\) value to \(h\) will not pass the HLT.
4. No, for instance \(f(1) = 3\) and \(f(0)= 3\) so there are two \(x\) values that give the same \(y\) value to \(f\) will not pass the HLT.
5. Yes, it is a function so we know it passes the VLT. It passes the HLT because of the restriction \(x\geq 4\).

Solution: The graph of \(h\) is determined by connecting points with straight lines. To graph the inverse we would need to use property 6 of inverses and connect the points Put it all together to get the graph of the inverse.

Solution:

1. \((f+g)(-3)=f(-3)+g(-3) = 4-5 = -1\)
2. \((f\cdot g)(2) = f(2)\cdot g(2) = 3\cdot 0 =0\)
3. \(\left(\dfrac{f}{g}\right)\!(-1) = \dfrac{f(-1)}{g(-1)}=\dfrac{2}{-3}\)
4. \((f\circ g)(3) = f(g(3))=f(1)=4\)
5. \(g^{-1}(-4) = -2\)
6. \((f\circ f)(2) = f(f(2))=f(3)=2\)
7. \(g(f(g(1)))=g(f(-1))=g(2)=0\)
8. The domain of \(f+g\) is \([-3,4]\)
9. The domain of \(\dfrac{f}{g}\) is \([-3,2)\cup(2,4]\)
10. \(g\) is a one-to-one function since it passes HLT and VLT.
11. \((f(3))^3-4g(-2) = (2)^3 - 4(-4) = 8+16=24\)
12. \(f(x)=3\) when \(x=-2, x=0, x=2\)