Another inverse function issue...

• Apr 20th 2009, 06:42 AM
struck
Another inverse function issue...
Given that f : x -> x^2 + 6x, and g: x -> 2x - 1 ... where x is a real number.

Find the two values of x such that fg(x) = gf(x), giving each answer in the form $\displaystyle p + q \sqrt{3}$

I have tried solving it by equating by solving it for both of the functions but that's not working. How should I proceed with it?
• Apr 20th 2009, 06:45 AM
Moo
Hello,
Quote:

Originally Posted by struck
Given that f : x -> x^2 + 6x, and g: x -> 2x - 1 ... where x is a real number.

Find the two values of x such that fg(x) = gf(x), giving each answer in the form $\displaystyle p + q \sqrt{3}$

Well...

$\displaystyle f(g(x))=(2x-1)^2+6(2x-1)=\text{expand everything}=4x^2+8x-5$

$\displaystyle g(f(x))=2(x^2+6x)-1=\dots=2x^2+12x-1$

Now solve for x in :
$\displaystyle 4x^2+8x-5=2x^2+12x-1$

that is $\displaystyle 2x^2-4x-4=0$

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

in order to do that, find the discriminant, and use your head... ^^
• Apr 20th 2009, 07:08 AM
struck
Sorry, but I should have posted the steps. I just wanted to know if I had to right idea. Ok, so far my steps were correct but at the last step, I was finding this using quadratic formula.

$\displaystyle \frac {2 \pm \sqrt{4 + 8}}{2}$

$\displaystyle \frac {2 \pm \sqrt{3} \sqrt {4}}{2}$

$\displaystyle \frac {2 \pm 2 \sqrt{3}}{2}$

The answer I am supposed to get is $\displaystyle 1 \pm \sqrt{3}$ but I am getting $\displaystyle 2 \pm \sqrt {3}$. I had, similarly, tried it without simplification at the last step ... can't seem to figure out what's wrong.
• Apr 20th 2009, 07:12 AM
Moo
Quote:

Originally Posted by struck
Sorry, but I should have posted the steps. I just wanted to know if I had to right idea. Ok, so far my steps were correct but at the last step, I was finding this using quadratic formula.

$\displaystyle \frac {2 \pm \sqrt{4 + 8}}{2}$

$\displaystyle \frac {2 \pm \sqrt{3} \sqrt {4}}{2}$

$\displaystyle \frac {2 \pm 2 \sqrt{3}}{2}$

The answer I am supposed to get is $\displaystyle 1 \pm \sqrt{3}$ but I am getting $\displaystyle 2 \pm \sqrt {3}$. I had, similarly, tried it without simplification at the last step ... can't seem to figure out what's wrong.

Yeah, by discriminant, I meant what you did.
And yes, it would have been better to show the steps :( avoiding some useless lines :(

Anyway, you've got the answer ! ;)

$\displaystyle \frac{2\pm 2 \sqrt{3}}{2} \neq 2 \pm \sqrt{3}$
I don't know how you got that !

$\displaystyle \frac{2 \pm 2 \sqrt{3}}{2}=\frac 22 \pm \frac{2 \sqrt{3}}{2}=1 \pm \sqrt{3}$ (Surprised)
• Apr 20th 2009, 07:23 AM
struck
OMG. My brain wasn't working well I guess. Thanks.