Pearson Algebra 2 Common Core, 2011
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Pearson Algebra 2 Common Core, 2011 View details
8. Graphing Radical Functions
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Exercise 65 Page 420

To determine the inverse of f(x), first replace f(x) with y. Then switch x and y and solve for y.

Inverse: f^(-1)(x) = (x+4)^2 - 3, x≥-4
Result: Yes, the inverse is a function.

Practice makes perfect
We will begin by finding the inverse of f(x). First, we need to replace f(x) with y. From there, we switch x and y and solve for y. y=sqrt(x+3)-4 → x=sqrt(y+3)-4 The resulting equation will be the inverse of the given function.
x=sqrt(y+3)-4
â–¼
Solve for y
x+4=sqrt(y+3)
(x+4)^2=y+3
y=(x+4)^2-3

Finally, to indicate that this is the inverse function of f(x), we will replace y with f^(- 1)(x). f^(- 1)(x)=(x+4)^2-3

Now that we found the inverse of f(x), we will determine the domain and range of f(x) and f^(-1).

Domain and Range of the Function

To determine the domain of the given function, the radicand cannot be negative. Let's find the value(s) of x that make this true.
x+3 ≥ 0
x≥-3
All values of x that are greater than or equal to -3 are included in the domain. Domain: x≥-3 For the range, think about the fact that the principal root of a number is always positive. lf(x) l= lsqrt(x+3) ≥0 l-4 Because the f(x) is difference of non-negative number and 4, the range is all values of y such that y≥-4. Range:& y≥-4

Domain and Range of the Inverse

When we find the inverse of a function, we are basically exchanging its x and y values. Therefore, the range of f(x) becomes the domain of f^(- 1)(x), and the domain of f(x) becomes the range of f^(- 1)(x). Domain: x ≥ -4 Range: y ≥ -3

Is the Inverse a Function?

A function is a relation where each input is related to exactly one output. In this case, for each x in domain of f^(- 1)(x), there is only one value of y in the range. Therefore, f^(- 1)(x) is a function in its domain.