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Math

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PROBLEM

Find f1(x)f^{-1}(x), and state the domain and range of f1(x)f^{-1}(x) for f(x)=12(x+5)f(x)=\sqrt{12}(x+5).

STEP 1

Assumptions1. The function is f(x)=12(x+5)f(x)=\sqrt{12}(x+5). We are looking for the inverse function, f1(x)f^{-1}(x)3. We also need to find the domain and range of f1(x)f^{-1}(x)

STEP 2

To find the inverse of a function, we switch xx and yy and then solve for yy. So, we start by replacing f(x)f(x) with yy.
y=12(x+5)y=\sqrt{12}(x+5)

STEP 3

Next, we swap xx and yy.
x=12(y+5)x=\sqrt{12}(y+5)

STEP 4

Now we solve for yy to find f1(x)f^{-1}(x). First, we divide both sides by 12\sqrt{12}.
x12=y+\frac{x}{\sqrt{12}}=y+

STEP 5

Then, we subtract5 from both sides to isolate yy.
y=x125y=\frac{x}{\sqrt{12}}-5

STEP 6

This is the expression for f1(x)f^{-1}(x).
f1(x)=x125f^{-1}(x)=\frac{x}{\sqrt{12}}-5

STEP 7

Now we need to find the domain and range of f1(x)f^{-1}(x). The domain of f1(x)f^{-1}(x) is the range of f(x)f(x), and the range of f1(x)f^{-1}(x) is the domain of f(x)f(x).

STEP 8

The domain of f(x)f(x) is all real numbers, because xx can be any real number in the function f(x)=12(x+5)f(x)=\sqrt{12}(x+5). Therefore, the range of f1(x)f^{-1}(x) is all real numbers.

SOLUTION

The range of f(x)f(x) is all real numbers greater than or equal to 512-5\sqrt{12}, because the smallest value f(x)f(x) can take is 512-5\sqrt{12} when x=x=. Therefore, the domain of f(x)f^{-}(x) is all real numbers greater than or equal to 512-5\sqrt{12}.
So, the inverse function is f(x)=x125f^{-}(x)=\frac{x}{\sqrt{12}}-5 with domain x512x \geq -5\sqrt{12} and range yy is all real numbers.

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