# A composite function problem

The question is: Suppose $$f(x) = x^2+1,$$ $$g(x) = 3-x.$$ Find the values for $$x$$ for such that $$(g\circ f)(x) = (f \circ g)(x).$$

I tried banging my head for one hour but my answer doesn't match the one given by the book which is $$1/\sqrt{2}$$ and $$-1/\sqrt{2}$$. I think the answer given in the book is wrong because I even tried putting the given answer in $$(g\circ f)(x)$$ and $$(f \circ g)(x)$$ and the two don't match up. My answer: $$x =\frac{3\pm\sqrt{-7}}{2}$$

• Do you know how to composite functions work?
– Gon
Jan 7 at 8:07
– Pedro
Jan 7 at 8:07
• Yeah, the answer definitely is not $\pm 1/\sqrt 2$ ($f$ would give a rational, $g$ would not). What answer did you get and how? We can look at that instead. Wouldn't be the first time a book has been wrong. Jan 7 at 8:20

$$f(g(x)) = g(x)^2 + 1 = (3-x)^2 + 1$$ $$g(f(x)) = 3 - f(x) = 3 - x^2 - 1$$

Hence we need to find the points $$x$$ that satisfy $$(3-x)^2 + 1 = 2 - x^2$$

We get that: $$9 -6x+ x^2 + 1 -2 + x^2 = 0 \implies$$ $$2x^2 -6x + 8 = 0 \implies$$ $$x^2 - 3x + 4 = 0 \implies$$ $$...$$

you are correct.

• How did you get $2x^2 - 6x + 8$? Jan 7 at 8:32
• @Ajay 9 + 1 -2 = 8. You made a mistake in your calculations Jan 7 at 8:36
• Whoops, I wrote the correct thing on paper but typed the wrong thing...ugghhh...Corrected now, thanks. Jan 7 at 8:39

Given that $$f(x) = x^2 + 1$$ and $$g(x) = 3 -x$$, the composite function should look as follows for the LHS of the equation, hence, $$3-(x^2 + 1)$$ RHS, $$(3-x)^2 + 1$$ Thus, $$3-(x^2 + 1) = (3-x)^2 + 1$$.

Simplifying we get, $$2x^2-6x+8 = 0\\ x^2 - 3x + 4 = 0.$$

In order to find the value's of $$x$$ satisfying the equation we can use the quadratic formula, and you should get $$\frac{3 \pm i\sqrt{7}}{2}$$ where $$i = \sqrt{-1}$$. Meaning you are correct.