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I have found the following problem.

If $x = -1$ and $x=2$ then $f(x) = 0$. If $g(x)= 2x-1$, then find the value of $x$ for which $f\circ g(x) = 0$.

I have solved the above problem in the following way.

$f(x) = (x+1)(x-2)$
$f\circ g(x) = f(g(x)) = (2x)(2x-3)$
So $2x(2x-3) = 0 \implies x= 0$ or $x= 3/2$
So if $x = 0$ or $x=3/2$ then $f\circ g(x) = 0$

Is the above procedure right? I have solved the above problem with a bit of confusion. That's why I am asking this question here. Please, remove my confusion with a proper explanation.

egreg
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Way to infinity
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2 Answers2

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Hint $f$ isn't necessary a polynomial.

Find the value of $x$ for which $g(x)=-1$ and $g(x)=2$.

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Yes: Assuming $f(x) = 0 $ if $x = -1,$ or $x = 2$, then one possible function for $f(x)$ will be

$f(x) = (x+1)(x-2)$.

And we are given $g(x) = 2x - 1$.

If this is the case, it follows that $(f\circ g)(x) = f(g(x)) = (2x)(2x-3)$

And so: $2x(2x-3) = 0\implies x= 0$ or $x= 3/2$

But, since $f(x)$ isn't necessarily a polynomial, or may be a multiple of our guess at $f(x)$, so we would need to check that that at these two values, it holds that $g(x) = -1$, or $g(x) = 2$.

Can you see why knowing precisely what $f(x)$ isn't necessary? (In fact, we don't know for certain what $f(x)$ actually is.)

Without knowing $f(x)$, but only knowing $f(-1) = 0$ and $f(2) = 0$, we can find the values at which $g(x) = -1,$ and when $g(x) = 2$, because then we know for sure that $$f(g(x)) =f(-1) = 0\quad\text{and that }\quad f(g(x)) = f(2) = 0$$

$$g(x) = 2x - 1 = -1 \iff 2x = 0 \iff \bf x = 0$$

$$g(x) = 2x - 1 = 2\iff 2x = 3 \iff \bf x =\dfrac 32$$

amWhy
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