Question

finding inverses
find an equation for the inverse for each of the following relations.

1. ( y = 3x + 2 )
2. ( y = -5x - 7 )
3. ( y = 12x - 3 )
4. ( y = -8x + 16 )
5. ( y = \frac{2}{3}x - 5 )
6. ( y = -\frac{3}{4}x + 5 )
7. ( y = -\frac{5}{8}x + 10 )
8. ( y = \frac{1}{2}x + 8 )
9. ( y = x^2 + 5 )
10. ( y = x^2 - 4 )
11. ( y = (x + 3)^2 )
12. ( y = (x - 6)^2 )

Explanation:

Let's solve for the inverse of the function \( y = 3x + 2 \) (problem 3) as an example.

Step 1: Swap x and y

To find the inverse, we first swap the roles of \( x \) and \( y \) in the equation. So we get \( x = 3y + 2 \).

Step 2: Solve for y

Now we solve this new equation for \( y \). Subtract 2 from both sides: \( x - 2 = 3y \). Then divide both sides by 3: \( y=\frac{x - 2}{3}\), which can be rewritten as \( y=\frac{1}{3}x-\frac{2}{3} \). So the inverse function \( f^{-1}(x)=\frac{1}{3}x - \frac{2}{3} \).

Let's do problem 4: \( y=- 5x - 7 \)

Step 1: Swap x and y

We have \( x=-5y - 7 \)

Step 2: Solve for y

Add 7 to both sides: \( x + 7=-5y \). Then divide both sides by - 5: \( y=-\frac{1}{5}x-\frac{7}{5} \), so the inverse function \( f^{-1}(x)=-\frac{1}{5}x-\frac{7}{5} \)

Problem 5: \( y = 12x-3 \)

Step 1: Swap x and y

\( x = 12y-3 \)

Step 2: Solve for y

Add 3 to both sides: \( x + 3=12y \). Divide both sides by 12: \( y=\frac{x + 3}{12}=\frac{1}{12}x+\frac{1}{4} \), so the inverse function \( f^{-1}(x)=\frac{1}{12}x+\frac{1}{4} \)

Problem 6: \( y=-8x + 16 \)

Step 1: Swap x and y

\( x=-8y + 16 \)

Step 2: Solve for y

Subtract 16 from both sides: \( x-16=-8y \). Divide both sides by - 8: \( y=-\frac{1}{8}x + 2 \), so the inverse function \( f^{-1}(x)=-\frac{1}{8}x + 2 \)

Problem 7: \( y=\frac{2}{3}x-5 \)

Step 1: Swap x and y

\( x=\frac{2}{3}y-5 \)

Step 2: Solve for y

Add 5 to both sides: \( x + 5=\frac{2}{3}y \). Multiply both sides by \( \frac{3}{2} \): \( y=\frac{3}{2}x+\frac{15}{2} \), so the inverse function \( f^{-1}(x)=\frac{3}{2}x+\frac{15}{2} \)

Problem 8: \( y =-\frac{3}{4}x + 5 \)

Step 1: Swap x and y

\( x=-\frac{3}{4}y + 5 \)

Step 2: Solve for y

Subtract 5 from both sides: \( x - 5=-\frac{3}{4}y \). Multiply both sides by \( -\frac{4}{3} \): \( y=-\frac{4}{3}x+\frac{20}{3} \), so the inverse function \( f^{-1}(x)=-\frac{4}{3}x+\frac{20}{3} \)

Problem 9: \( y =-\frac{5}{8}x + 10 \)

Step 1: Swap x and y

\( x=-\frac{5}{8}y + 10 \)

Step 2: Solve for y

Subtract 10 from both sides: \( x - 10=-\frac{5}{8}y \). Multiply both sides by \( -\frac{8}{5} \): \( y=-\frac{8}{5}x + 16 \), so the inverse function \( f^{-1}(x)=-\frac{8}{5}x + 16 \)

Problem 10: \( y=\frac{1}{2}x + 8 \)

Step 1: Swap x and y

\( x=\frac{1}{2}y + 8 \)

Step 2: Solve for y

Subtract 8 from both sides: \( x - 8=\frac{1}{2}y \). Multiply both sides by 2: \( y = 2x-16 \), so the inverse function \( f^{-1}(x)=2x - 16 \)

Problem 11: \( y=x^{2}+5 \)
First, we need to note that the function \( y = x^{2}+5 \) is not one - to - one over the entire real number line. But if we consider the domain \( x\geq0 \) (or \( x\leq0 \)):

Step 1: Swap x and y

\( x=y^{2}+5 \)

Step 2: Solve for y

Subtract 5 from both sides: \( x - 5=y^{2} \). Then \( y=\pm\sqrt{x - 5} \). If we consider the domain \( x\geq0 \) for the original function, the range of the original function is \( y\geq5 \), and the inverse function (for \( x\geq5 \)) is \( y = \sqrt{x - 5} \) (if we took the domain \( x\leq0 \) for the original function, the inverse function would be \( y=-\sqrt{x - 5} \) for \( x\geq5 \))

Problem 12: \( y=x^{2}-4 \)
The function \( y=x^{2}-4 \) is not one - to - one over \( \mathbb{R} \). For domain \( x\geq0 \) (or \( x\leq0 \)):

Step 1: Swap x and y

\( x=y^{2}-4 \)

Step 2: Solve for y

Add 4 to both sides: \( x + 4=y^{2} \). Then \( y=\pm\sqrt{x + 4} \). If the original function has domain \( x\geq0 \), range \( y\geq - 4 \), and the inverse function (for \( x\geq - 4 \)) is \( y=\sqrt{x + 4} \) (for \( x\geq0 \) original domain) or \( y =-\sqrt{x + 4} \) (for \( x\leq0 \) original domain)

Problem 13: \( y=…

Answer:

s (for the linear functions, the inverses are):

1. \( f^{-1}(x)=\frac{1}{3}x-\frac{2}{3} \)
2. \( f^{-1}(x)=-\frac{1}{5}x-\frac{7}{5} \)
3. \( f^{-1}(x)=\frac{1}{12}x+\frac{1}{4} \)
4. \( f^{-1}(x)=-\frac{1}{8}x + 2 \)
5. \( f^{-1}(x)=\frac{3}{2}x+\frac{15}{2} \)
6. \( f^{-1}(x)=-\frac{4}{3}x+\frac{20}{3} \)
7. \( f^{-1}(x)=-\frac{8}{5}x + 16 \)
8. \( f^{-1}(x)=2x-16 \)

For the quadratic functions (11 - 14), the inverses are defined piece - wise (depending on the domain of the original function):

1. If domain of \( y=x^{2}+5 \) is \( x\geq0 \), \( f^{-1}(x)=\sqrt{x - 5},x\geq5 \); if domain is \( x\leq0 \), \( f^{-1}(x)=-\sqrt{x - 5},x\geq5 \)
2. If domain of \( y=x^{2}-4 \) is \( x\geq0 \), \( f^{-1}(x)=\sqrt{x + 4},x\geq - 4 \); if domain is \( x\leq0 \), \( f^{-1}(x)=-\sqrt{x + 4},x\geq - 4 \)
3. If domain of \( y=(x + 3)^{2} \) is \( x\geq - 3 \), \( f^{-1}(x)=-3+\sqrt{x},x\geq0 \); if domain is \( x\leq - 3 \), \( f^{-1}(x)=-3-\sqrt{x},x\geq0 \)
4. If domain of \( y=(x - 6)^{2} \) is \( x\geq6 \), \( f^{-1}(x)=6+\sqrt{x},x\geq0 \); if domain is \( x\leq6 \), \( f^{-1}(x)=6-\sqrt{x},x\geq0 \)