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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: $$ x =\frac{3\pm\sqrt{-7}}{2} $$

s01itarius 1upus
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  • Do you know how to composite functions work? – Gon Jan 07 '22 at 08:07
  • Please improve your title to something more informative. – Pedro Jan 07 '22 at 08:07
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    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. – PrincessEev Jan 07 '22 at 08:20

2 Answers2

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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.

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

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.

JustANoob
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