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If $f(x) = 2x + 1$ and $g(x) = x + 3$, then which of the following is true?
(A) $(f + g)x = f(x) + g(x)$
(B) $(f \circ g)(x) = (g \circ f)(x)$
(C) $(fg)(x) = g[f(x)]$
(D) $(fg)(x) = f[g(x)]$

I came across this question while studying "Functions". I computed and confirmed that (B) is false.

I found out from google that options (A), (C), (D) are "Operations of Functions" and that $(f+g)(x)$ is defined as $f(x)+g(x)$ (like operating on polynomials). So, option (A) should be true.

The given answer is option (A). Why are options (C) and (D) false?

InanimateBeing
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  • "From the theory, I know that (B) is false." -- that is correct in the general case, but (B) could still be true for special cases, like f(x) = 2x, g(x) = 3x. So you have to check for that. – g.kertesz Aug 22 '22 at 13:20
  • @g.kertesz right! I don't think there's any theory like that. Edited it to what I actually did. (And I'm quite sure that that's the (only?) way to do it) – InanimateBeing Aug 22 '22 at 14:08

2 Answers2

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Usually, if $f,g: I \to \mathbb{R}$ are real-valued functions defined on some non-empty set $I$, then we define: $$(fg)(x) := f(x)g(x)$$ There are some contexts in which $fg$ can refer to composition of both functions but we won't worry about them now. In the context of the math you're working on now, they're probably going with the definition I stated above. So, it follows that $C$ and $D$ are false.

Abhijeet Vats
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(A) $(f + g)x = f(x) + g(x)$

Yes, as you have pointed out, this is true by definition.

(B) $(f \circ g)(x) = (g \circ f)(x)$
(C) $(fg)(x) = g[f(x)]$
(D) $(fg)(x) = f[g(x)]$

We can infer from Mordeus's explanation that options B, C, D are false for the given $f$ and $g.$

However, it's worth noting that these three equations are not identically false: they are true for $f(x):=g(x):=1,$ for example.

Oh! I got confused between $f∘g(x)$ and $fg(x).$

$$f\circ g(x)\equiv fg(x)\equiv f(g(x))\not\equiv(fg)(x).$$

ryang
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