Is $f(g(x))$ discontinuous?

Question

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?

Answer 1

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.

Answer 2

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 $\{(x,y)\in\mathbb{R^2} : x\neq \frac{1}{10}(15+\sqrt{5}),x\neq \frac{3}{2}-\frac{1}{2\sqrt5}\}$.