Problem figuring out inverse function (looks easy but I'm confused)

• Oct 31st 2009, 09:18 AM
s3a
Problem figuring out inverse function (looks easy but I'm confused)
Normally when you try to find f-1(x) you just switch the x and the y for example:

f(x) = x+3
x = y + 3
y = x - 3
f-1(x) = x- 3

but in the following problem:
"The formula C = 5/9(F-32), where F>= - 459.67, expresses the Celsius temperature C as a function of the Fahrenheit temperature F. Find a formula for the inverse function and interpret it. What is the domain of the inverse function?"

The answer for the formula is F = 9/5*C + 32 which I also obtain if I solve for F (Note NOT switch the places of x (in this case F) and y (in this case C). My question is: why is this the case?

Could someone please explain the reasoning to me?
• Oct 31st 2009, 09:49 AM
tonio
Quote:

Originally Posted by s3a
Normally when you try to find f-1(x) you just switch the x and the y

No, you don't. Instead of expressing y as a function of x you do the other way around: express x as a function of y. This is not "switching places" between x, y, otherwise you'd get for the inverse function EXACTLY the same expression as for the original function, which is not the case...and, btw, which is NOT what you did: you did the correct thing.

for example:

f(x) = x+3
x = y + 3
y = x - 3
f-1(x) = x- 3

but in the following problem:
"The formula C = 5/9(F-32), where F>= - 459.67, expresses the Celsius temperature C as a function of the Fahrenheit temperature F. Find a formula for the inverse function and interpret it. What is the domain of the inverse function?"

The answer for the formula is F = 9/5*C + 32 which I also obtain if I solve for F (Note NOT switch the places of x (in this case F) and y (in this case C). My question is: why is this the case?

Could someone please explain the reasoning to me?

You ALSO didn't "switch" places in the example you gave, otherwise you'd get $\displaystyle f(x)= y = x+3 \Longleftrightarrow x=y+3$, which is false since the inverse function actually is $\displaystyle x=y-3\Longrightarrow f^{-1}(x)=x-3$, and thus the same's here: if you have $\displaystyle C=\frac{5}{9}(F-32)$, then expressing as function of C you get precisely $\displaystyle C=\frac{5}{9}(F-32)\Longleftrightarrow \frac{9}{5}\,C=F-32\Longleftrightarrow F=\frac{9}{5}\,C+32$ , and we're done.

Tonio
• Oct 31st 2009, 11:11 AM
s3a
Wait something is wrong my example:
you said that I'm right but according to your explanation, I am wrong also in my example.
"Look":
I got f-1(x) = x - 3 = y
and you said x = y - 3 --> f-1(x) = x - 3 but if you rearrange what you wrote in bold, you'd get y = x+3 and not y = x-3.

I did switch in my example because:
f(x) = y and f-1(x) = y (of inverse function)
y = x + 3
x = y + 3
y = x - 3
f-1(x) = x - 3

Something is either wrong or really confusing here so I was hoping I could get this cleared before proceeding to the actual problem.
• Oct 31st 2009, 02:40 PM
tonio
Quote:

Originally Posted by s3a
Wait something is wrong my example:
you said that I'm right but according to your explanation, I am wrong also in my example.
"Look":
I got f-1(x) = x - 3 = y
and you said x = y - 3 --> f-1(x) = x - 3 but if you rearrange what you wrote in bold, you'd get y = x+3 and not y = x-3.

It doesn't matter how you call your variables: if $\displaystyle f(x)=x+3$ then the inverse function is $\displaystyle g(x)=x-3$ because $\displaystyle f(g(x))=g(f(x))=x$

I did switch in my example because:
f(x) = y and f-1(x) = y (of inverse function)
y = x + 3
x = y + 3
y = x - 3
f-1(x) = x - 3

Something is either wrong or really confusing here so I was hoping I could get this cleared before proceeding to the actual problem.

Again: to find the inverse function of y in terms of x, you must write x in terms of y, and then you can switch (Hey! This time it is switch) the variables for simplicity.
Anyway, and in case case of doubt, just compose the original function with the one you thinbk is its inverse: if you get the identity function you're fine, otherwise you're not.

Tonio
• Oct 31st 2009, 04:35 PM
HallsofIvy
Tonio, you can do it either way and most people are taught to swap variable first.:

"Late swapping", f(x)= y= x+ 3 so y- 3= x. Now swap: $\displaystyle f^{-1}(x)= y= x- 3$.

"Early swapping", f(x)= y= x+ 3 so swapping variables, x= y+ 3. Now solve for y: x- 3= y so $\displaystyle f^{-1}(x)= y= x-3$.

I prefer "early swapping" because I think the crucial point to make is that going from f to f inverse does swap domain and range. The rest is just manipulation!

The "Fahrenheit", "Celcius" is just an annoying abberation! If you had f(x)= 5/9(x-32) and were asked to find the inverse function, you would say "y= (5/9)(x- 32) so I first swap x and y: x= (5/9)(y- 32) and solve for y: (9/5)x= y- 32 so y= (9/5)x+ 32. $\displaystyle f^{-1}(x)= (9/5)x+ 32$.

However, here, the variables have specific physical meaning. In the first x was temperature in degrees Fahrenheit, which we call "F" to remind us of that and the result of the function is the temperature in degrees Celcius which we call "C".

Now, the inverse function is $\displaystyle f^{-1}(x)$= (9/5)x+ 32 but now we are going the other way (that's what "inverse" does) so to remind ourselves of this, we Label "$\displaystyle f^{-1}$ as "F" and x as "C".
F= (9/5)C+ 32.