SOLUTION: Find the inverse function of f:[3,∞) -> R, f(x) = (x-4)^2 +1 Why does the domain of f need to be restricted to [4,∞) in this question?

Algebra ->  Vectors -> SOLUTION: Find the inverse function of f:[3,∞) -> R, f(x) = (x-4)^2 +1 Why does the domain of f need to be restricted to [4,∞) in this question?      Log On


   

Question 1028867: Find the inverse function of f:[3,∞) -> R, f(x) = (x-4)^2 +1
Why does the domain of f need to be restricted to [4,∞) in this question?
Found 2 solutions by jim_thompson5910, josgarithmetic:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Here is the basic outline for finding the inverse
A) Replace f(x) with y (step 2 in table below)
B) Swap x and y (step 3 in table below)
C) Solve for y (steps 4 through 7 in table below)
Let's use this to find the inverse of the given function
NumberStatementReason/Explanation
1.f%28x%29+=+%28x-4%29%5E2+%2B1None needed. This is the original function given
2.y+=+%28x-4%29%5E2+%2B1Replace f(x) with y
3.x+=+%28y-4%29%5E2+%2B1Swap x and y. Now we isolate y.
4.x-1+=+%28y-4%29%5E2Subtract 1 from both sides
5.%28y-4%29%5E2+=+x-1 Flip the equation
6.y-4+=+sqrt%28x-1%29Take the square root of both sides. See note below
7.y=+sqrt%28x-1%29%2B4+Add 4 to both sides


Note: the domain of [4,infinity) for f(x) turns out to be the range of the inverse function. Domain of original = range of inverse. In order to stretch onto positive infinity, we need to use the plus version of the plus/minus. So instead of using plus/minus, we can just use plus all by itself. So instead of using y-4+=+%22%22%2B-sqrt%28x-1%29 we stick with y-4+=+sqrt%28x-1%29

------------------------------------------------------------------------------

The inverse function is therefore

The domain of f(x) must be restricted to [4,infinity) because including 3 in the domain makes the function not one-to-one. Notice how f(3) = f(5) = 2. You must restrict the domain to make f(x) one-to-one in order for f(x) to be invertible to a function.

Answer by josgarithmetic(39618) About Me  (Show Source):
You can put this solution on YOUR website!
Look first for the inverse relation to f.

%28R%28x%29-4%29%5E2%2B1=x for the inverse function of f.
%28R%28x%29-4%29%5E2=x-1
R%28x%29-4=0%2B-+sqrt%28x-1%29
R%28x%29=4%2B-+sqrt%28x-1%29

The domain of R(x) is for TWO different functions. R itself is a relation and not a function. This is because the "plus or minus" part of the expression. The domain for either branch of R is x%3E=4.

Look again at function f(x). Domain is ALL REAL NUMBERS. What about the range of f(x)? f(x) is a parabola with a vertex minimum value at (4,1). This means that the RANGE for f(x) is x%3E=4, or as a description, from 4 inclusive toward infinity.

Going from a function (f(x)) to its inverse, the domain and range switch. The range x%3E=4 for f(x) is the DOMAIN for either branch of R(x).



Function f(x)
graph%28300%2C300%2C-3%2C12%2C-3%2C12%2C%28x-4%29%5E2%2B1%29

Upper branch of R(x)
graph%28300%2C300%2C-3%2C12%2C-7%2C8%2C-4%2Bsqrt%28x-1%29%29

Lower branch of R(x)
graph%28300%2C300%2C-3%2C12%2C-7%2C8%2C-4-sqrt%28x-1%29%29