C3 range and domain help

Does anyone know how to get to the final answer when finding the inverse of a function? I always seem to get it right but its never the way the mark scheme wants it, and the mark scheme doesnt seem to allow my version, even though its equivalent.

Any tips on how to get it the way the mark scheme wants it? Thanks

(btw if it helps the question is g=1/(2X+1). Find g^-1)

Reply 1

muffinmachine

10

Replace the Xs with Ys and vice versa, then rearrange for Y

In this case

y=\frac{1}{2x+1} \to x= \frac{1}{2y+1}

\frac{1}{x}=2y+1

y = \frac{1}{2x}-\frac{1}{2}

g^{-1}(x) = \frac{1}{2x}-\frac{1}{2}

(edited 5 years ago)

Reply 2

muffinmachine

10

Just thought I'd add, this isn't really about the range or domain of functions - range and domain are the y and x values a function can take, respectively.

(edited 5 years ago)

Reply 3

Prasiortle

13

Original post by SomethingQuarky

Does anyone know how to get to the final answer when finding the inverse of a function? I always seem to get it right but its never the way the mark scheme wants it, and the mark scheme doesnt seem to allow my version, even though its equivalent.

Any tips on how to get it the way the mark scheme wants it? Thanks

(btw if it helps the question is g=1/(2X+1). Find g^-1)

Please could you give an example of "my version" and "the way the mark scheme wants it", so I can see what you're talking about. In general, they will allow equivalent forms unless the question requires you to give the answer in a particular form.

C3 range and domain help

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