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The inverse of a function

f(x)= 9/x+3
Find F-1(x) This was a question on one of my maths papers for uni.
I get x=6 via
Which I can see is wrong when I plot my original function on a graph .
Can anyone answer this. What really bothers me is that it appears to be simple and probably is. :colondollar::mad::shot:
Reply 1
Avatar for Notnek
Notnek
21
Original post by JakeSuCeng
f(x)= 9/x+3
Find F-1(x) This was a question on one of my maths papers for uni.
I get x=6 via
Which I can see is wrong when I plot my original function on a graph .
Can anyone answer this. What really bothers me is that it appears to be simple and probably is. :colondollar::mad::shot:

Is it 9x+3 \displaystyle \frac{9}{x} + 3 \ or  9x+3\ \displaystyle \frac{9}{x+3}?

You need to use brackets. Also, what do you mean by "via"?
Reply 2
Avatar for JakeSuCeng
JakeSuCeng
OP
Original post by notnek
Is it 9x+3 \displaystyle \frac{9}{x} + 3 \ or  9x+3\ \displaystyle \frac{9}{x+3}?

You need to use brackets. Also, what do you mean by "via"?


The second one oh its not actually in brackets on the paper but it appears like that.
Original post by JakeSuCeng
f(x)= 9/x+3
Find F-1(x) This was a question on one of my maths papers for uni.
I get x=6 via
Which I can see is wrong when I plot my original function on a graph .
Can anyone answer this. What really bothers me is that it appears to be simple and probably is. :colondollar::mad::shot:


x=9f−1(x)+3⇒f−1(x)=3(3−x)xx = \dfrac{9}{f^{-1}(x)+3} \Rightarrow f^{-1}(x) = \dfrac{3(3-x)}{x}

How did you get x=6?x = 6?
(edited 9 years ago)
Reply 4
Avatar for Notnek
Notnek
21
Original post by JakeSuCeng
The second one oh its not actually in brackets on the paper but it appears like that.

It doesn't need to have brackets if it's clear. But you need to use brackets when posting on this forum to make it clear.

To find the inverse, rearrange to make x the subject e.g.

f(x)=x+3f(x) = x+3

x=f(x)−3x = f(x) - 3

So the inverse of f(x)f(x) is f−1(x)=x−3\displaystyle f^{-1}(x)=x-3
Reply 5
Avatar for JakeSuCeng
JakeSuCeng
OP
Original post by CTArsenal
x=9f−1(x)+3⇒f−1(x)=3(3−x)xx = \dfrac{9}{f^{-1}(x)+3} \Rightarrow f^{-1}(x) = \dfrac{3(3-x)}{x}

How did you get x=6?x = 6?

By
y= 9/x+3 (*x+3)
xy +3y =9
xy=9-3y (/y)
x=6
Reply 6
Avatar for Notnek
Notnek
21
Original post by JakeSuCeng
By
y= 9/x+3 (*x+3)
xy +3y =9
xy=9-3y (/y)
x=6

Your last line is wrong. When you divide by y, you have to divide everything on the right by y which includes 9 and -3y.
Reply 7
Avatar for JakeSuCeng
JakeSuCeng
OP
okay so now I got f-1(x) = (9/x) -3
Reply 8
Avatar for Notnek
Notnek
21
Original post by JakeSuCeng
okay so now I got f-1(x) = (9/x) -3

That's correct.
This is considered undergraduate? wow what has happened to maths at uni these days. Than god i am going to imperial though :rolleyes: proper maths there.
Original post by Infamous7
...


Don't be condescending. This math may well be for a course other than Mathematics. Also, don't feel so high and mighty about going to Imperial. I bet I could pick a whole load of students from any other university that'd wipe the floor with your "mathematical ability".
Reply 11
Avatar for Notnek
Notnek
21
Original post by Infamous7
This is considered undergraduate? wow what has happened to maths at uni these days. Than god i am going to imperial though :rolleyes: proper maths there.

How childish.

And how can you know what's "happened to maths at uni these days" without yet experiencing uni?

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