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Given $f(x)=\dfrac{1}{x-1}$ and $g(x)=\dfrac{1}{x-2}$, it is obvious that both are having infinite discontinuities at $x=1$ and $x=2$. Now for$$(f \circ g)(x)=\frac{x-2}{3-x},$$ where is it discontinuous? No doubt at $x=3$, but why $(f \circ g)(x)$ is also discontinuous at $x=2$?

Ѕᴀᴀᴅ
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Ekaveera Gouribhatla
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    Because it is not defined at $x=2$ – gammatester Jun 23 '18 at 08:07
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    The domain of the composite function is $x\in \mathbb{D}(g)\cap g(x)\in \mathbb{D}(f)$. So it is not defined at $x=2$ –  Jun 23 '18 at 08:07
  • I don't know how this problem was formulated, nor how "infinite discontinuities" was defined, but the concept of continuity requires an element of the domain. It makes as much sense to ask if $f\circ g$ is continuous at $2$ as it does asking if it continuous at $\begin{bmatrix}1 & 1\ 1 & 0 \end{bmatrix}$ or at "shoe". My point is that $f\circ g$ is not continuous at $2$ and it is not discontinuous at $2$, the pseudo statement has no meaning. – Git Gud Jun 23 '18 at 08:52

3 Answers3

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By definition, ($f$ $\circ$ $g$)($x$) = $f(g(x))$. So ($f$ $\circ$ $g$)$(2)$ = $f(g(2))$. But $g$ is discontinuous at $x=2$, so $f \circ g$ is not defined for $x=2$, so it's not continuous.

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You have $f(x)=\frac{1}{x-1}$ and $g(x)=\frac{1}{x-2}$ ; so

$f(g(x))=\frac{1}{g(x)-1} = \frac{1}{\frac{1}{x-2}-1}=\frac{x-2}{3-x}$ ;

Consider the second equation for your answer If (g(x)) would not exist at all at (x=2) then where would you find (f(g(x)) to compute?

Arijit Dey
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To answer your second question: $f \circ g$ means that you first apply $g$ and then apply $f.$ You cannot apply $g$ to $x=2$, so $f \circ g$ is not defined at $x=2.$

Pawel
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  • Yes I agree $fog$ is not defined at $x=2$ but limit of $fog(x)$ from both left and right is zero. So it's discontinuous – Ekaveera Gouribhatla Jun 23 '18 at 08:48
  • The first condition for the function to be continuous at $x$ is that the function has to be defined at $x.$ Since your function is not defined at 2, it is not continuous at $2$. The limit is irrelevant in this case. – Pawel Jun 23 '18 at 13:26