Maths Help. (1 Viewer)

[Dylanamali]

f(x) = 2x + Inx, find f^-1(2). Give reasons.

 

[math man]

ok now with inverse functions all x and y values swap, so finding f ^ -1 (2) is the same as saying 2= 2x + lnx, which obviously occurs when x=1.
so therfore f^-1(2)=1

 

[math man]

if you still are unsure, i will go through an example:
let x=2, i.e y(2) = 4 + ln2, so the point (4+ln2, 2) is on our original function, now this point corresponds to (2, 4 + ln2) on the inverse function as i just
swap x and y values as above. So your question f^-1(2) is basically the point (2, y_1) on your inverse graph which becomes ( y_1,2) on the original fucntion
which means y_1 =1 as above

 

[Dylanamali]

what is the inverse?

so firstly you let f(x) = y,
therefore y = 2x + Inx
then to find inverse, you swap x and y:
x = 2y + Iny

how do you solve that equation for y?

 

[math man]

you can't solve that, and in 2u you're not meant, you just have to understand the swapping of x and y values to do this question

 

[Dylanamali]

what is y_1?

 

[Dylanamali]

I understand that if you firstly just do f(2) you get y=4+In2, so therefore on the graph f(x) there's a point (2, 4+In2).
And then therefore on the inverse graph f-1(x) there's a point (4+In2, 2).
Now where do you go after that?

 

[Dylanamali]

and i'm not sure if this is 2u.
This is UNI maths, but I think should be on the basis of 2u/3u maths.

 

[Carrotsticks]

how do you solve that equation for y?

To your knowledge, there is no real way of solving it algebraically, but we can get a solution by inspection (x=1).

what is y_1?

Think about it this way:

When dealing with the inverse, the X values become Y values.

We want the Y value of f^(-1)(x) when x=2, so we will do so by finding the X value of f(x) when y=2, which is x=1 by inspection. They are the same thing.

f^(-1)(2) --> y = ???? = f(2) = x = ??? = 1

Hence f^(-1)(2) --> y = 1

 

[Dylanamali]

ahh yeah as i was reading it over it made sense.. thank you to both of you. repped+