# Thread: Functions - Pre Calculus

1. ## Functions - Pre Calculus

Functions f and g are defined as follows
$\displaystyle f(x) = 2x + 3, x\epsilon\Re$ and $\displaystyle g(x) = \frac{1}{x-1}, x\epsilon\Re$, $\displaystyle x$$\displaystyle \ne 1 (i) Find the expression for the inverse function f^-1\displaystyle (x) (ii) Find the expression for the composite function \displaystyle gf$$\displaystyle (x)$

(iii) Solve the equation f^-1$\displaystyle (x)$ $\displaystyle =$$\displaystyle gf(x)-1 I got: (i) \displaystyle y = 2x+3 \displaystyle y - 3 = 2x \displaystyle \frac{y-3}{2} = x \displaystyle \frac{x-3}{2} = y ? (ii) And I got \displaystyle g(f(x)) = \frac{1}{2x+2} Am i correct? If not, please help. 2. Correct on both questions, what did you get for the third? 3. I can't do to be honest. I can only plug in the values to the equation. If anyone could post the solution, I would be grateful. 4. You already know that \displaystyle f^{-1}(x)= \frac{x-3}{2} and that \displaystyle g(f(x))= \frac{1}{2x+2}. To solve \displaystyle f^{-1}(x)= g(f(x)), set \displaystyle \frac{x-3}{2}= \frac{1}{2x+ 2} and solve for x. I recommend multiplying both sides by 2x+2= 2(x+1) to get rid of the fractions. That will result in a quadratic equation. 5. Could you post the solution? I have similar ones to these and I could work from that one if you posted it. I stuggle with this kind of math. 6. Originally Posted by gary223 Could you post the solution? I have similar ones to these and I could work from that one if you posted it. I stuggle with this kind of math. I'm happy to give you a kick start. \displaystyle \frac{x-3}{2}= \frac{1}{2x+ 2} \displaystyle (x-3)(2x+2)= 2 \displaystyle 2x^2-4x-6=2 \displaystyle 2x^2-4x-8=0 The rest is yours to finish here's a hint: The Quadratic Formula Explained 7. Shouldn't there be a '-1' after the \displaystyle \frac{1}{2x+2} since its in part (iii) of the question? Or is it how you posted? 8. Originally Posted by gary223 (iii) Solve the equation f^-1\displaystyle (x) \displaystyle =$$\displaystyle gf(x)-1$
It seems there should be,

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

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,

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

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

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

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

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

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

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

10. I have

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

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

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

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

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

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

continue...

11. Originally Posted by pickslides
I have

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

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

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

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

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

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

continue...

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

$\displaystyle 2x^2+2x-2x-2=2$

$\displaystyle 2x^2-2=2$

$\displaystyle 2x^2 = 0$

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

$\displaystyle 2x^2+2x-2x-2=2$

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

$\displaystyle 2x^2 = 4$

13. Thank you.