# Find the values of $k$ for which the equation $(f\circ g)(x) = x$ has two equal roots

I'm busy doing a problem which asks the above considering the following:

$$f(x) = 4x - 2k\text{ and }g(x) = 9/(2-x)$$

As far as I know roots usually refer to quadratics and even when doing the composition I still can't figure out how to find the two equal roots.

After the composition I get the following:

$$(f\circ g)(x) = (36/(2-x)) - 2k$$

Now, I don't know how I am suppose to turn that into a quadratic to find out the roots, unless the problem is just semantics and the question is really asking for the intercept and not the roots.

Thank you!

Bernard

• You need to set your composite expression equal to $x$ (as stated in the title), after multiplying through by $(2-x)$ the resulting equation will be quadratic in $x$. – okrzysik Feb 19 '16 at 10:45

You said that

$$(f\circ g)(x)=\frac{36}{2-x}-2k$$

But as both you and okrzysik pointed out, the composition $(f\circ g)(x)=x$ and so

$$(f\circ g)(x)=\frac{36}{2-x}-2k=x$$

So multiplying both sides by $2-x$ yields

$$(2-x)x=36-2k(2-x)\Rightarrow x^2+(2k-2)x+(36-4k)$$

So the solution by QF is

$$x=\frac{2-2k\pm\sqrt{4k^2-8k+4-4(36-4k)}}{2}$$ $$=1-k\pm\sqrt{k^2+2k-35}$$

For the problem at hand, we need the two roots to be equal, so

$$1-k+\sqrt{k^2+2k-35}=1-k-\sqrt{k^2+2k-35}$$

$$\sqrt{k^2+2k-35}=-\sqrt{k^2+2k-35}$$

$$k^2+2k-35=0 \Rightarrow (k+7)(k-5)=0...$$

and solve for $k$ from here.... But, to check to see if this is valid or not, realize that if $(f\circ g)(x)=x$, then $g=f^{-1}$ and so

$$f^{-1}(x)=\frac{x+2k}{4}$$

and since $g=f^{-1}$, we get that

$$\frac{x+2k}{4}=\frac{9}{2-x}$$

Solving by cross multiplication yields the same quadratic as above.

• I would just like to clarify, under "So the solution by QF is", should the 8k not be -8k? – Bernard Feb 20 '16 at 8:31
• Hand slapping face, yes @BernardFrankel, – Eleven-Eleven Feb 20 '16 at 12:39