Question

1. find the inverse of the relation for parts a - e.

a. $f(x) = \sqrt3{x} + 6$
b. $f(x) = 5x^2 + 8$
c. $f(x) = \frac{2}{3}x + 7$
d. $f(x) = \sqrt{5x + 1}$
e. $f(x) = \frac{x + 1}{2}$

Explanation:

Part a:

Step1: Replace \( f(x) \) with \( y \)

\( y = \sqrt[3]{x} + 6 \)

Step2: Swap \( x \) and \( y \)

\( x = \sqrt[3]{y} + 6 \)

Step3: Solve for \( y \)

Subtract 6 from both sides: \( x - 6 = \sqrt[3]{y} \)
Cube both sides: \( y = (x - 6)^3 \)

Step4: Replace \( y \) with \( f^{-1}(x) \)

\( f^{-1}(x) = (x - 6)^3 \)

Part b:

Step1: Replace \( f(x) \) with \( y \)

\( y = 5x^2 + 8 \)

Step2: Swap \( x \) and \( y \)

\( x = 5y^2 + 8 \)

Step3: Solve for \( y \)

Subtract 8: \( x - 8 = 5y^2 \)
Divide by 5: \( \frac{x - 8}{5} = y^2 \)
Take square roots: \( y = \pm\sqrt{\frac{x - 8}{5}} \) (Note: The original handwritten answer had a sign error and incorrect numerator; the correct inverse for \( y = 5x^2 + 8 \) (which is not one - to - one over all real numbers, but the inverse relation is \( f^{-1}(x)=\pm\sqrt{\frac{x - 8}{5}} \))

Step4: Replace \( y \) with \( f^{-1}(x) \)

\( f^{-1}(x)=\pm\sqrt{\frac{x - 8}{5}} \)

Part c:

Step1: Replace \( f(x) \) with \( y \)

\( y=\frac{2}{3}x + 7 \)

Step2: Swap \( x \) and \( y \)

\( x=\frac{2}{3}y + 7 \)

Step3: Solve for \( y \)

Subtract 7: \( x - 7=\frac{2}{3}y \)
Multiply both sides by \( \frac{3}{2} \): \( y=\frac{3}{2}(x - 7)=\frac{3}{2}x-\frac{21}{2} \)

Step4: Replace \( y \) with \( f^{-1}(x) \)

\( f^{-1}(x)=\frac{3}{2}x-\frac{21}{2} \)

Part d:

Step1: Replace \( f(x) \) with \( y \)

\( y = \sqrt{5x + 1} \) (assuming \( y\geq0 \) for the principal square root)

Step2: Swap \( x \) and \( y \)

\( x=\sqrt{5y + 1} \)

Step3: Solve for \( y \)

Square both sides: \( x^{2}=5y + 1 \)
Subtract 1: \( x^{2}-1 = 5y \)
Divide by 5: \( y=\frac{x^{2}-1}{5} \) (with the domain restriction \( x\geq0 \) for the inverse to be a function)

Step4: Replace \( y \) with \( f^{-1}(x) \)

\( f^{-1}(x)=\frac{x^{2}-1}{5},x\geq0 \)

Part e:

Step1: Replace \( f(x) \) with \( y \)

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

Step2: Swap \( x \) and \( y \)

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

Step3: Solve for \( y \)

Multiply both sides by 2: \( 2x=y + 1 \)
Subtract 1: \( y = 2x-1 \)

Step4: Replace \( y \) with \( f^{-1}(x) \)

\( f^{-1}(x)=2x - 1 \)

Answer:

s:
a. \( \boldsymbol{f^{-1}(x)=(x - 6)^3} \)

b. \( \boldsymbol{f^{-1}(x)=\pm\sqrt{\frac{x - 8}{5}}} \)

c. \( \boldsymbol{f^{-1}(x)=\frac{3}{2}x-\frac{21}{2}} \)

d. \( \boldsymbol{f^{-1}(x)=\frac{x^{2}-1}{5}(x\geq0)} \)

e. \( \boldsymbol{f^{-1}(x)=2x - 1} \)