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Question: Let $f(x) = \frac{1}{15x^2+8x+1} $ and $ g(x)= \frac{1}{(x-1)(x-2)} $, then the number of points of discontinuity of $f(g(x))$ is?

The answer key claimed that the answer is $1$, but I don't agree with it. I claim that the answer should be zero, since $f(g(x))$ simplifies to $h(x)=\frac{x^{2}-3x+2}{(5x^{2}-15x+11)(3x^{2}-9x+7)}$. So $h(1)=h(2)=0$, which upon plotting on Desmos gives rise to same results.

However, there is a counter argument that since $g(x)$ is not defined at $x=1,2$, how can $f(g(x))$ be defined at those points? And hence there are $2$ points of discontinuity.

Which of the arguments is correct?

Rodrigo de Azevedo
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mrtechtroid
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    To define a function, it is necessary to define its domain and codomain. What are the domains and codomains of your functions f and g ? The points of discontinuity of a function must be points of its domain. – Dan Asimov May 23 '23 at 17:51
  • @DanAsimov the original question did "NOT" specify its domain or codomain for both f(x),g(x) . – mrtechtroid May 23 '23 at 17:52
  • It should be at least two points as you can't calculate $g(x)$ at $x=1, 2$ – Vasili May 23 '23 at 17:53
  • There is also a convention, that your book may be following, which is to assume the domain of any real-valued function is the set of x for which the function's definition makes sense as a real number. – Dan Asimov May 23 '23 at 17:54
  • @DanAsimov So you mean both 0 and 2 are valid results just in different domains? (in $R$ and $R-{1,2}$ respectively for g(x) – mrtechtroid May 23 '23 at 17:56
  • Domain of $f(g(x))$ is the intersection of domain $g(x)$ and domain of the resulting composition. If domain $g(x)$ is not specified, it's hard to tell what the author of the problem meant. – Vasili May 23 '23 at 18:01
  • mrtechnoid — what I mean is that x = 1 and x = 2 cannot possibly be in the domain of the function g. – Dan Asimov May 23 '23 at 18:05
  • Perhaps not relevant: your $h(x)$ has $4$ complex points for which the denominator would be zero – Henry May 23 '23 at 18:54
  • [[1]] ZERO is not right. [[2]] It is not defined at 2 Points 1,2 + Denominator of the new function has maximum 4 roots (2 in this Case) hence 2+4=6 is the maximum number of Points of Discontinuity. In this Case , that is 2+2=4 Points of Discontinuity. [[3]] ONE is not right too. – Prem May 23 '23 at 19:10
  • You're composing two continuous functions, so the result must be continuous. See also this: https://math.stackexchange.com/questions/1087623/is-function-f-mathbb-c-0-rightarrow-mathbb-c-prescribed-by-z-rightarrow – Hans Lundmark May 23 '23 at 19:30
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    If it's not apparent already from the comments, you should have included your acting definition of "$f$ is discontinuous at $a$" or equivalently "$a$ is a point of discontinuity of $f$" to preclude all this confusion. – Brian Moehring May 23 '23 at 19:48
  • @Vasili: $\mathsf{dom} (f \circ g) = g^{-1}[\mathsf{dom}(f)]$. That isn't quite the same as what you wrote. – Rob Arthan May 23 '23 at 20:18
  • The number $0$ is correct (based on the definitions of discontinuity I'm familiar with), but as Paul Frost points out in his answer, your reasoning isn't quite on target. This question is related. – Cameron Buie May 24 '23 at 01:09

2 Answers2

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It is important to observe that a function can only be discontinuous in points where it is defined, i.e. in points of its domain.

In your question no domains of $f$ and $g$ are specified.

Let $Z(f) = \{ -\frac 1 5, - \frac 1 3 \}$ and $Z(g) = \{ 1, 2 \}$ denote the set of zeros of the denominators $p(x) = 15x^2 + 8x + 1$ of $f$ and $q(x) = (x-1)(x-2) = x^2 -3x -2$ of $g$. Therefore the maximal domains for $f$ and $g$ are $D(f) = \mathbb R \setminus Z(f) = \mathbb R \setminus \{ -\frac 1 5, - \frac 1 3 \}$ and $D(g) = \mathbb R \setminus Z(g) = \mathbb R \setminus \{ 1, 2 \}$. Both $f$ and $g$ are continuous on their maximal domains.

If we additionally assign function values to $-\frac 1 5$ and $-\frac 1 3$, we get an extended function $\bar f$ with domain $\mathbb R$, but whatever our choice of function values was, $\bar f$ is discontinuous in the points $-\frac 1 5, - \frac 1 3$. A similar assertion holds for $g$. However, the question does not contain any hint that we should work with such extensions $\bar f$ and $\bar g$.

It is easy to verify that $g$ does not attain the values $-\frac 1 5, - \frac 1 3$ (which means that $q$ does not attain the values $-5, -3$). Hence the range of $g$ is contained in the domain of $f$. The composition $f \circ g$ therefore has domain $D(g)$ and is continuous on it (each composition of continuous functions is continuous).

This shows that the number of points of discontinuity of $f \circ g$ is zero.

You say that $f(g(x))$ simplifies to $h(x) = \frac{x^{2}-3x+2}{(5x^{2}-15x+11)(3x^{2}-9x+7)}$. This is not true since $h(0) = \frac{2}{77}$ and $f(g(0)) = f(\frac 1 2) = \frac{4}{35}$.

Actually we have $$f(g(x)) = \frac{1}{\frac{15}{q(x)^2} + \frac{8}{q(x)} +1} .$$ For $x \ne 1, 2$ (which is equivalent to $q(x) \ne 0$) we get $$\frac{1}{\frac{15}{q(x)^2} + \frac{8}{q(x)} +1} = \frac{q(x)^2}{15 + 8q(x) +q(x)^2} =:h(x).$$ $h(x)$ is also defined for $x = 1,2$ and has the value $0$ at these points. Thus $f \circ g$ has $h$ as a continuous extension to $\mathbb R$. But be aware that the functions $f \circ g$ and $h$ are not the same since their domains do not agree.

Paul Frost
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You first of all must define a domain for the functions in order to determine if these functions are continuos or not. Clearly all three functions all not continuous on $\mathbb{R}$, namely there are points of infinite discontinuity. We can remove these points to make these functions continuous. For h(x)=$\frac{x^{2}-3x+2}{(5x^{2}-15x+11)(3x^{2}-9x+7)}$ these are the points with $x=\frac{3}{2}-\frac{1}{2\sqrt5}$ and $x=\frac{1}{10}(15+\sqrt5)$. So $h(x)$ is continuous on $\{(x,y)\in\mathbb{R^2} : x\neq \frac{1}{10}(15+\sqrt{5}),x\neq \frac{3}{2}-\frac{1}{2\sqrt5}\}$.

muhammed gunes
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