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I do not understand the first example right after the topic of function composition is introduced in Mathematik für Ingenieure by Thomas Reissinger.

There is a function $g : \mathbb{R} \rightarrow [0, \infty)$ defined as

$$g(x) = x^2 + 1$$

and a function $f : [0, \infty) \rightarrow \mathbb{R}$ defined as

$$f(x) = \sqrt x$$

in order to explain $h = f \circ g$.

I understand that $[0, \infty)$ would be appropriate as domain for $f$. But why is the codomain of $g$ $[0, \infty)$ and not $[1, \infty)$?

no1dea
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1 Answers1

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Codomain is not the same as the image of the function.

The codomain of a function is a set into which all of the output is constrained to fall (blue oval below). It's like a dartboard.

The image of a function is the set of all output values it produces (yellow oval below). It's like the region on the dartboard in which the darts from the domain will actually land.

enter image description here So while the codomain of $g$ is given here as $[0,\infty),$ its image is $g(\mathbb{R})=[1,\infty).$

Likewise, while the codomain of $f$ is given here as $\mathbb{R},$ its image is $f([0,\infty))=[0,\infty).$

Golden_Ratio
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    Fond of illustrations! – Mikasa Mar 18 '23 at 08:27
  • Thank you very much! So would it be appropriate to say that the codomain has to be defined but the image can be determined? – no1dea Mar 18 '23 at 08:29
  • @no1dea The codomain is defined, but keep in mind it must defined to be large enough to contain the image. The dartboard has to be large enough to contain all the places darts will land! – Golden_Ratio Mar 18 '23 at 08:34
  • @Golden_Ratio I got it, thank you! It all makes sense regarding codomain and image but the only thing that is still puzzling me is why did the author select such a confusing example? I personally think that $g(x) = x^2$ would have been much more clear. – no1dea Mar 18 '23 at 08:40
  • @no1dea My guess is the author felt it would more clearly illustrate composition of two functions. Defining $g(x)=x^2$ would render $h$ simplified as $h(x)=|x|$; perhaps that might confuse introductory students learning about composition? – Golden_Ratio Mar 18 '23 at 08:45
  • @Golden_Ratio Could be, yes. Anyhow, I've learnt something new today, thanks a lot for your help! – no1dea Mar 18 '23 at 08:54
  • @no1dea no prob! – Golden_Ratio Mar 18 '23 at 08:55