Functions

so this is multi step C3 question on Functions:

f(x)= x^2 -6 , x E R , X>0
find inverse f(x)
so i found it to be Root (X+6), X E R, X>-6

it then asked me to plot the graphs and i did so.
then in part c) it asked me to find X when f(x)=inverse f(x)
so i established that X^2-6= Root(X+6) how do i solve this?

Hope people can decode that lol i dunno how to use the maths symbols on here.
X E R ( X in a real set of numbers)

Reply 1

anatomical frog

Well start by getting rid of that square root, and try to get everything on one side. Then the rest should follow (assuming I've read that right).

(edited 10 years ago)

Reply 2

m4ths/maths247

Original post by cuckoo99

When you drew the two graphs what line did you reflect f(x) in?
What do you think that tells you about the solution to the equation?

Reply 3

m4ths/maths247

Original post by anatomical frog

Well start by getting rid of that square root, and try to get everything on one side. Then the rest should follow (assuming I've read that right).

It isn't required when you look at it graphically and set two different equations equals. :smile:

Reply 4

cuckoo99

Original post by anatomical frog

Well start by getting rid of that square root, and try to get everything on one side. Then the rest should follow (assuming I've read that right).

if i square both sides i get (X^2-6)^2= x+6
so no that isn't the full story

Reply 5

anatomical frog

Original post by m4ths/maths247

It isn't required when you look at it graphically and set two different equations equals. :smile:

Oh of course! Sorry, I'm a bit tired of the maths after all my exams! :colondollar:

Reply 6

anatomical frog

Original post by cuckoo99

if i square both sides i get (X^2-6)^2= x+6
so no that isn't the full story

And although the above poster has the better idea, you could continue by expanding the brackets, then doing the algebraic rearranging...

Reply 7

cuckoo99

Original post by m4ths/maths247

When you drew the two graphs what line did you reflect f(x) in?
What do you think that tells you about the solution to the equation?

judging from my graph it looks like the line of reflection is y=x lol but my question was more aimed at solving f(x)=f^-1(x)

Edit: only learnt this on friday so ye im rusty to say the least

(edited 10 years ago)

Reply 8

m4ths/maths247

Original post by cuckoo99

judging from my graph it looks like the line of reflection is y=x lol but my question was more aimed at solving f(x)=f^-1(x)

My post is aimed at solving the equation you have asked about.

Have a think about it logically.

Reply 9

cuckoo99

Original post by m4ths/maths247

My post is aimed at solving the equation you have asked about.

Have a think about it logically.

When i drew the graph i used the domain given to me to find the range of the f(x) to then find the domain of the inverse function. i know the point of intersection will be my value of X the problem is my drawing sucks ass :biggrin:

i think the point of intersection is (3,3) so without doing the horrible algebra i supose X = 3

Edit: Checked the answer and it is indeed 3 :biggrin:

thanks i need to learn to draw better graphs lol im so sloppy with it

(edited 10 years ago)

Reply 10

m4ths/maths247

Original post by cuckoo99

i think the point of intersection is (3,3) so without doing the horrible algebra i supose X = 3

Edit: Checked the answer and it is indeed 3 :biggrin:

thanks i need to learn to draw better graphs lol im so sloppy with it

I think often one can fall blindly into following an algorithm on these topics.
The graphs are reflected in the line y = x
If y = x meets y = x^2 - 6 then x = x^2 -6 and you have a straight forward simultaneous equation to deal with and one solution to pick (Which should be obvious from the graphs).
You could have used the other equation of course and set that = x. :smile: