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allegansveritatem

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My goal is to find the inverse of the function. I am trying to isolate the x. Here is what I get--I have done this numerous times and get the same result and it is NOT the inverse of the original:
inverse.PNG

What is the problem here? I know it is wrong because I have checked it both graphically and by plugging in some values of the original function. So...what gives?
 
You are almost finished. Instead of \(\displaystyle x= \frac{\frac0{8y+ 5}{y}}{3- \frac{4}{y}}\), a fraction over a fraction, simplify by multiplying both numerator and denominator by y: \(\displaystyle x= \frac{8y+ 5}{3y- 4}\).

(I honestly can't see why you chose to divide both sides by y in your sixth line. From your fifth line, \(\displaystyle 8y+ 5= 3xy- 4x\), factor x out of the right side to get \(\displaystyle 8y+ 5= x(3y- 4)\) and then divide both sides by 3y- 4: \(\displaystyle x= \frac{8y+ 5}{3y- 4}\).)
 
allegansveritatem said:
My goal is to find the inverse of the function. I am trying to isolate the x. Here is what I get--I have done this numerous times and get the same result and it is NOT the inverse of the original:
View attachment 14033

What is the problem here? I know it is wrong because I have checked it both graphically and by plugging in some values of the original function. So...what gives?
Click to expand...
4th line from bottom - we have:

8y + 5 = 3xy - 4x = (3y - 4)* x

x = (8y + 5) /(3y - 4)

Your mistake is not in "inverse function" calculation. You are making mistake while checking your work!

Check at:

https://www.wolframalpha.com/input/?i=Inverse+function+of+y+=+(4*x+5)/(3*x-8)
 
HallsofIvy said:
You are almost finished. Instead of \(\displaystyle x= \frac{\frac0{8y+ 5}{y}}{3- \frac{4}{y}}\), a fraction over a fraction, simplify by multiplying both numerator and denominator by y: \(\displaystyle x= \frac{8y+ 5}{3y- 4}\).

(I honestly can't see why you chose to divide both sides by y in your sixth line. From your fifth line, \(\displaystyle 8y+ 5= 3xy- 4x\), factor x out of the right side to get \(\displaystyle 8y+ 5= x(3y- 4)\) and then divide both sides by 3y- 4: \(\displaystyle x= \frac{8y+ 5}{3y- 4}\).)
Click to expand...
I like that idea in the parentheses. That looks like the way to go...but it is late and I am frazzled so I will have to go over this tomorrow. I will post my results.
One thing: You have me posting something I don't recognize, namely:fraction 1.PNG
Is that a typo? Or is it some transformation of my result? It doesn't look like it is a viable expression. I mean, a numerator of zero...como?
 
Subhotosh Khan said:
4th line from bottom - we have:

8y + 5 = 3xy - 4x = 3y - 4)* x

x = (8y + 5) /(3y - 4)

Your mistake is not in "inverse function" calculation. You are making mistake while checking your work!

Check at:

https://www.wolframalpha.com/input/?i=Inverse+function+of+y+=+(4*x+5)/(3*x-8)
Click to expand...
I will look at this tomorrow when my brain comes back from where it goes every night about this time. I will say that I still don't get why what I posted is not working. I worked this out again earlier this evening and got something that does produce an inverse effect. It was still pretty homely as functions go but it worked. I will photo it tomorrow and post it.
 
so, I went back at it today and quickly, with the help of suggestions above, I came to a fairly elegant inverse:

inverse2.PNG

But I still don't know why the first inverse--the one with the wacky bunch of fractions--didn't work. Seems that it should. I mean, it may be unsightly but looks aren't everything.
 
Your work is fine.
Why do you think your original work is wrong? Have you plugged in any particular values for y into your two solutions and got back different x values. I do admit that your original solution does not allow y=0.
 
allegansveritatem said:
What is the problem here? I know it is wrong because I have checked it both graphically and by plugging in some values of the original function. So...what gives?
Click to expand...
allegansveritatem said:
But I still don't know why the first inverse--the one with the wacky bunch of fractions--didn't work. Seems that it should. I mean, it may be unsightly but looks aren't everything.
Click to expand...
What exactly happened when you did those checks? Please show us.

Apart from the fact that your original form excludes y=0, as was pointed out, it's easy to convert it to the nicer form, just by multiplying numerator and denominator by y, as has been said already. So the checks should both succeed, if you did them correctly.

Of course, the proper, full check is to substitute your expression for x into the original function and see that you get y back out.
 
Dr Peterson,
Multiplying the OP's answer by y (without looking back at where it came from) will not allow y to take on the value 0.
 
True, the first method shown was a detour that messed up the domain of the inverse, so it's not the best way to have done it. That's not my point.

My point is merely that that error doesn't explain how the checks would have failed, unless the OP only tried y=0. That's what I'm interested in at the moment: that the answer was wrong only on a technicality and should have passed the test.
 
Dr.Peterson said:
What exactly happened when you did those checks? Please show us.

Apart from the fact that your original form excludes y=0, as was pointed out, it's easy to convert it to the nicer form, just by multiplying numerator and denominator by y, as has been said already. So the checks should both succeed, if you did them correctly.

Of course, the proper, full check is to substitute your expression for x into the original function and see that you get y back out.
Click to expand...
Well, I can only say that when I checked it yesterday, I got poor results but today it is OK. I think I may have entered it wrong into my calculator and got a graph back that was linear.
Yes, the big drawback is no zero.
On another note, there is one thing I learned: If the rational function is composed of two linear expressions the inverse is easy to get: All you do is take the coefficient on the variable in the numerator, change the sign, and put it in place of the constant. Then, take that displaced constant, change the sign, and let it be the new coefficient in the numerator and Hey, Presto! you have the inverse. I stumbled on this yesterday. I'm hoping to make a fortune on it.
 
Jomo said:
Your work is fine.
Why do you think your original work is wrong? Have you plugged in any particular values for y into your two solutions and got back different x values. I do admit that your original solution does not allow y=0.
Click to expand...
I think I must have been to tired to do a good job of testing...and I think I entered it wrong into my graphing calculator to boot. But it passes now! Thanks. I like the second one better though, it doesn't exclude the zero.
 
allegansveritatem said:
Well, I can only say that when I checked it yesterday, I got poor results but today it is OK. I think I may have entered it wrong into my calculator and got a graph back that was linear.
Yes, the big drawback is no zero.
On another note, there is one thing I learned: If the rational function is composed of two linear expressions the inverse is easy to get: All you do is take the coefficient on the variable in the numerator, change the sign, and put it in place of the constant. Then, take that displaced constant, change the sign, and let it be the new coefficient in the numerator and Hey, Presto! you have the inverse. I stumbled on this yesterday. I'm hoping to make a fortune on it.
Click to expand...
Right! That's why mathematicians are all wealthy!
 
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