# Thread: Functions - Pre Calculus

1. ## Functions - Pre Calculus

Functions f and g are defined as follows
$
f(x) = 2x + 3, x\epsilon\Re$
and $g(x) = \frac{1}{x-1}, x\epsilon\Re$, $x$ $\ne 1$

(i) Find the expression for the inverse function f^-1 $(x)$

(ii) Find the expression for the composite function $gf$ $(x)$

(iii) Solve the equation f^-1 $(x)$ $=$ $gf(x)-1$

I got:
(i)
$y = 2x+3$

$y - 3 = 2x$

$\frac{y-3}{2} = x$

$\frac{x-3}{2} = y$?

(ii)
And I got $g(f(x)) = \frac{1}{2x+2}$

2. Correct on both questions, what did you get for the third?

3. I can't do to be honest. I can only plug in the values to the equation. If anyone could post the solution, I would be grateful.

4. You already know that $f^{-1}(x)= \frac{x-3}{2}$ and that $g(f(x))= \frac{1}{2x+2}$.

To solve $f^{-1}(x)= g(f(x))$, set $\frac{x-3}{2}= \frac{1}{2x+ 2}$ and solve for x.

I recommend multiplying both sides by 2x+2= 2(x+1) to get rid of the fractions. That will result in a quadratic equation.

5. Could you post the solution? I have similar ones to these and I could work from that one if you posted it. I stuggle with this kind of math.

6. Originally Posted by gary223
Could you post the solution? I have similar ones to these and I could work from that one if you posted it. I stuggle with this kind of math.
I'm happy to give you a kick start.

$\frac{x-3}{2}= \frac{1}{2x+ 2}$

$(x-3)(2x+2)= 2$

$2x^2-4x-6=2$

$2x^2-4x-8=0$

The rest is yours to finish

here's a hint: The Quadratic Formula Explained

7. Shouldn't there be a '-1' after the $\frac{1}{2x+2}$ since its in part (iii) of the question? Or is it how you posted?

8. Originally Posted by gary223
(iii) Solve the equation f^-1 $(x)$ $=$ $gf(x)-1$
It seems there should be,

$\frac{x-3}{2}= \frac{1}{2x+ 2}-1$

This thread has given you all the tools you need to solve this equation. Look forward to seeing your answer.

9. Originally Posted by pickslides
It seems there should be,

$\frac{x-3}{2}= \frac{1}{2x+ 2}-1$

This thread has given you all the tools you need to solve this equation. Look forward to seeing your answer.
Okay, I got:

$2x^2-4x-5=0$

$(2x+1)(x-5)=0$

$2x+1= 0$ and $x-5=0$

$2x=-1$ and $x=5$

$x=-\frac{1}{2}$ and $x=5$

10. I have

$\frac{x-3}{2}= \frac{1}{2x+ 2}-1$

$\frac{x-3}{2}+1= \frac{1}{2x+ 2}$

$\frac{x-3}{2}+\frac{2}{2}= \frac{1}{2x+ 2}$

$\frac{x-3+2}{2}= \frac{1}{2x+ 2}$

$\frac{x-1}{2}= \frac{1}{2x+ 2}$

$(x-1)(2x+2)= 2$

continue...

11. Originally Posted by pickslides
I have

$\frac{x-3}{2}= \frac{1}{2x+ 2}-1$

$\frac{x-3}{2}+1= \frac{1}{2x+ 2}$

$\frac{x-3}{2}+\frac{2}{2}= \frac{1}{2x+ 2}$

$\frac{x-3+2}{2}= \frac{1}{2x+ 2}$

$\frac{x-1}{2}= \frac{1}{2x+ 2}$

$(x-1)(2x+2)= 2$

continue...

$(x-1)(2x+2)= 2$

$2x^2+2x-2x-2=2$

$2x^2-2=2$

$2x^2 = 0$

12. Originally Posted by gary223
$(x-1)(2x+2)= 2$

$2x^2+2x-2x-2=2$

$2x^2-2=2$
Adding 2 to both sides gives

$2x^2 = 4$

13. Thank you.