Question

find the inverse of each function.
a. $y = \frac{x - 6}{4}$
b. $f(x) = x^2 + 9$
c. $f(x) = \sqrt{x - 3}$

Explanation:

Part a: \( y = \frac{x - 6}{4} \)

Step 1: Swap \( x \) and \( y \)

To find the inverse, we first interchange \( x \) and \( y \) in the equation. So we get:
\( x=\frac{y - 6}{4} \)

Step 2: Solve for \( y \)

Multiply both sides of the equation by \( 4 \) to get rid of the denominator on the right - hand side:
\( 4x=y - 6 \)
Then, add \( 6 \) to both sides of the equation to solve for \( y \):
\( y = 4x+6 \)
The inverse function is \( f^{-1}(x)=4x + 6 \)

Part b: \( f(x)=x^{2}+9 \)

Step 1: Replace \( f(x) \) with \( y \)

We have \( y=x^{2}+9 \)

Step 2: Swap \( x \) and \( y \)

Interchanging \( x \) and \( y \), we obtain \( x=y^{2}+9 \)

Step 3: Solve for \( y \)

Subtract \( 9 \) from both sides:
\( x - 9=y^{2} \)
Take the square root of both sides. Since the original function \( y = x^{2}+9 \) is not one - to - one (it fails the horizontal line test, for example, \( f(1)=1 + 9=10 \) and \( f(- 1)=1 + 9 = 10\)), the inverse is not a function unless we restrict the domain of the original function. If we consider the domain of \( f(x)=x^{2}+9 \) as \( x\geq0 \), then \( y=\sqrt{x - 9} \); if we consider the domain as \( x\leq0 \), then \( y=-\sqrt{x - 9} \). In general, the relation inverse to \( y=x^{2}+9 \) is \( y=\pm\sqrt{x - 9} \), but if we assume the function is defined for \( x\geq0 \) (to make it one - to - one), the inverse function is \( f^{-1}(x)=\sqrt{x - 9},x\geq9 \)

Part c: \( f(x)=\sqrt{x - 3} \)

Step 1: Replace \( f(x) \) with \( y \)

We have \( y = \sqrt{x - 3} \)

Step 2: Swap \( x \) and \( y \)

Interchanging \( x \) and \( y \), we get \( x=\sqrt{y - 3} \)

Step 3: Solve for \( y \)

Square both sides of the equation to eliminate the square root:
\( x^{2}=y - 3 \)
Then, add \( 3 \) to both sides of the equation to solve for \( y \):
\( y=x^{2}+3 \)
We need to consider the domain of the original function. The domain of \( f(x)=\sqrt{x - 3} \) is \( x\geq3 \), and the range is \( y\geq0 \). For the inverse function, the domain will be \( x\geq0 \) and the range will be \( y\geq3 \). So the inverse function is \( f^{-1}(x)=x^{2}+3,x\geq0 \)

Answer:

s:
a. The inverse function is \( \boldsymbol{f^{-1}(x)=4x + 6} \)

b. If the domain of \( f(x)=x^{2}+9 \) is restricted to \( x\geq0 \), the inverse function is \( \boldsymbol{f^{-1}(x)=\sqrt{x - 9},x\geq9} \); if the domain is restricted to \( x\leq0 \), the inverse function is \( \boldsymbol{f^{-1}(x)=-\sqrt{x - 9},x\geq9} \). In general, the relation inverse to \( y = x^{2}+9 \) is \( \boldsymbol{y=\pm\sqrt{x - 9}} \)

c. The inverse function (with the appropriate domain restriction \( x\geq0 \)) is \( \boldsymbol{f^{-1}(x)=x^{2}+3,x\geq0} \)