Thread: HELP! completely confused: inverse function and its domain and range for f(x)=3x+4

1. HELP! completely confused: inverse function and its domain and range for f(x)=3x+4

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.

2. Let's begin with:

$y=3x+4$

Now, to find the inverse, switch $x$ and $y$, and solve for $y$:

$x=3y+4$

$3y=x-4$

$\displaystyle y=\frac{x-4}{3}$

You were close, you just had a sign wrong.

And so, given:

$f(x)=3x+4$

We've found:

$\displaystyle f^{-1}(x)=\frac{x-4}{3}$

Let's check our work:

$\displaystyle f\left(f^{-1}(x)\right)= 3\left(\frac{x-4}{3}\right)+4= x-4+4=x\quad\checkmark$

So, we know we've found the inverse correctly.

A linear function of the form:

$f(x)=mx+b$ where $m\ne0$

will have all reals as the domain and all reals as the range. In other words, a non-horizontal and non-vertical line will be defined for all real $x$, and will be unbounded on either end.

I don't know what that list of numbers is suppose to represent...normally a domain/range is given in interval or set notation.

3. Originally Posted by melisasandoval
...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.
When you composed the function and its inverse (which should get a different name, such as f-1(x) or g(x), to avoid confusion) (here), what did you get? If you got "x", then your inverse is correct.

Originally Posted by melisasandoval
My teacher also gives us the answers but he didn't put the inverse functions just the domain and range.
I'm sorry, but I don't understand what you mean by the above...?

Originally Posted by melisasandoval
I plugged in the domain...
What do you mean by "plugging in a domain"? Into what did you "plug" it?

Originally Posted by melisasandoval
...but I don't get anywhere near his ranges.
What did you get? (We can't check work we can't see, to please show all of your steps.) What were "his ranges"? Did he get more than one range for the given function and domain?

Originally Posted by melisasandoval
The domain is 0,1,0,2.
For what are these three x-values the domain?

Originally Posted by melisasandoval
I also thought that a function couldn't have 0 twice as an x value or it wouldn't be a function.
Where are you seeing that zero (or any specific value) is included twice as an x-value?

Originally Posted by melisasandoval
Can someone please explain this? Thank you.
By definition, the domain of the original function is the range of the inverse function, and the range of the original function is the domain of the inverse function. (last example here) Find the domains of each function; this will give you the ranges of their counterparts.

4. A linear function, such as y= 3x+ 5, has "all numbers" as its "natural domain" (the set of all values of x for which we can evaluate 3x+ 5) but it is possible to define a new function by giving a formula and assigning a domain to it.

For example, I can define f(x)= 3x+ 5 with domain [0, 5]. That is the same formula as above but now I am asserting that it is only defined for $0\le x\le 5$. Since that is linear and increasing and f(0)= 5, f(5)= 20, the function's range is [5, 20]. Now, if asked for the inverse function, I would not only "reverse" the formula to $f^{-1}(x)= \frac{x- 5}{3}$ but I would also have to reverse the domain and range: the inverse function is $f^{-1}(x)= \frac{x- 5}{3}$ with domain [5, 20] and then the range is [0, 5].

5. Originally Posted by melisasandoval
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.
$f(f^{-1}(x)) = x = f^{-1}(f(x)).$

You can always use this to check your work.

$f(x) = 3x + 4\ and\ f^{-1}(x) = \dfrac{x + 4}{3} \implies$

$f(f^{-1}(x)) = f \left ( \dfrac{x + 4}{3} \right ) = 3 * \dfrac{x + 4}{3} + 4 = x + 4 + 4 = x + 8 \ne x.$

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.

$f(x) = 3x + 4$

$f(x) + 4 = 3x$ ERROR This of course is a simple error that may happen when you (or anyone) get distracted.

$x = \dfrac{f(x) + 4}{3}$

$f^{-1}(x) = \dfrac{x + 4}{3}$

Now let's avoid the error.

$f(x) = 3x + 4$

$f(x) - 4 = 3x$

$x = \dfrac{f(x) - 4}{3}$

$f^{-1}(x) = \dfrac{x - 4}{3}$

Let's check.

$f(f^{-1}(x)) = f \left ( \dfrac{x - 4}{3} \right ) = 3 * \dfrac{x - 4}{3} + 4 = x - 4 + 4 = x.$

That looks good. Let's try the other way.

$f^{-1}f(x)) = f^{-1} (3x + 4) = \dfrac{(3x + 4) - 4}{3} = \dfrac{ 3x}{ 3} = x.$

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.

Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•