Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 8.2
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Exercise 77 Page 393

Practice makes perfect
a Before we can find the inverse of the given function, we need to replace f(x) with y.
f(x)=2x-3 ⇔ y=2x-3 Now, to algebraically determine the inverse of the equation we exchange x and y and solve for y. Given Equation & Inverse Equation y=2 x-3 & x=2 y-3 The result of isolating y in the new equation will be the inverse of the given function.
x=2y-3
â–¼
Solve for y
x+3 = 2y
x+3/2 = y
y= x+3/2
Finally, we write the inverse of the given function in function notation by replacing y with f^(- 1)(x) in our new equation. f^(- 1)(x)=x+3/2
b To find the inverse of the given function, we again need to replace h(x) with y. From there, we will switch x and y and solve for y.
y=( x-3)^2+2 ↓ x=( y-3)^2+2The resulting equation will be the inverse of the given function.
x=(y-3)^2+2
â–¼
Solve for y
x-2 = (y-3)^2
sqrt(x-2)=sqrt((y-3)^2)

sqrt(a^2)=± a

±sqrt(x-2)=y-3
y-3=±sqrt(x-2)
y = ±sqrt(x-2)+3
Finally, to indicate that this is the inverse function of h(x), we replace y with h^(- 1)(x). h^(- 1)(x)= ± sqrt(x-2) + 3 Note that h^(- 1)(x) is not a function. By taking the square root, we created a positive value and a negative value. This gives two outputs for each input, and so by definition it is not a function.