HI. So I have been stuck on this question for about an hour now and I still don't know what to do. My teacher is teaching us about Inverse Functions and I'm puzzled. The question is about finding the inverse function and its domain and range for f(x)=3x+4. I did the inverse function but I probably did it wrong. I got f(x)=4/3+x/3. My teacher also gives us the answers but he didn't put the inverse functions just the domain and range. I plugged in the domain but I don't get anywhere near his ranges. The domain is 0,1,0,2. I also thought that a function couldn't have 0 twice as an x value or it wouldn't be a function. Can someone please explain this? Thank you.
\(\displaystyle f(f^{-1}(x)) = x = f^{-1}(f(x)).\)
You can always use this to check your work.
\(\displaystyle f(x) = 3x + 4\ and\ f^{-1}(x) = \dfrac{x + 4}{3} \implies\)
\(\displaystyle f(f^{-1}(x)) = f \left ( \dfrac{x + 4}{3} \right ) = 3 * \dfrac{x + 4}{3} + 4 = x + 4 + 4 = x + 8 \ne x.\)
So you
DID get the wrong answer, but you can also see pretty easily what the right answer should be.
Now you did not tell us how you arrived at your almost correct answer so we cannot be sure where you made your error, but it probably arose like this.
\(\displaystyle f(x) = 3x + 4\)
\(\displaystyle f(x) + 4 = 3x\)
ERROR This of course is a simple error that may happen when you (or anyone) get distracted.
\(\displaystyle x = \dfrac{f(x) + 4}{3}\)
\(\displaystyle f^{-1}(x) = \dfrac{x + 4}{3}\)
Now let's avoid the error.
\(\displaystyle f(x) = 3x + 4\)
\(\displaystyle f(x) - 4 = 3x\)
\(\displaystyle x = \dfrac{f(x) - 4}{3}\)
\(\displaystyle f^{-1}(x) = \dfrac{x - 4}{3}\)
Let's check.
\(\displaystyle f(f^{-1}(x)) = f \left ( \dfrac{x - 4}{3} \right ) = 3 * \dfrac{x - 4}{3} + 4 = x - 4 + 4 = x.\)
That looks good. Let's try the other way.
\(\displaystyle f^{-1}f(x)) = f^{-1} (3x + 4) = \dfrac{(3x + 4) - 4}{3} = \dfrac{ 3x}{ 3} = x.\)
That answer checks. So now you know how to check your work.
In the future, please show your work so we can point out where and why you made any errors.