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# Quick Help with Functions please C3 watch

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1. Nvm
2. (Original post by MrToodles4)
f(x) = 2+x^1/2

part 1) find an expression for f^-1(x) - I got this: f^-1(X) = (x-2)^2

part 2). Sketch the graph in the same set of axes the graph of f(x0 and the graph of f^-1(x), clearly marking the line of reflection between the two graphs. - I also got this.

Part 3). Show that x=4 is the only solution of the equation f(x) = f^-1(x) - This is the bit I do not understand - normally I just take the original function so int his case 2+x^1/2 and equal this to y=x and solve - but the answer says you cannot use this method - I don't understand why so any help will be greatly appreciated. They took f^-1(x) and equalled this to y=x, Why is this the case?! won't both lines (all 3 points) intersect at y=x???
What do you mean equate to y=x?

Solve f^-1 (x) = f (x), upon solving you square both sides which causes extra solutions which you have to check in the original equation to see if they hold true.
3. (Original post by NotNotBatman)
What do you mean equate to y=x?

Solve f^-1 (x) = f (x), upon solving you square both sides which causes extra solutions which you have to check in the original equation to see if they hold true.
I got the answer now thank you. I got X = 4 and X =1. The answer says x=4 since x>2 (I think thats the domain of the new f^-1(x) function)
4. (Original post by MrToodles4)
I got the answer now thank you. I got X = 4 and X =1. The answer says x=4 since x>2 (I think thats the domain of the new f^-1(x) function)
It's because 1 doesn't work in the original equation
5. (Original post by NotNotBatman)
It's because 1 doesn't work in the original equation
Oh I see - so stuff like this I need to be weary of. Thanks

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