Find two discontinuous functions whose product is continuous

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Hint: Find functions $f(x)$ and $g(x)$ such that their product $f(x)g(x)$ equals the constant function $0$ but where $f(x)$ is discontinuous at $x = 0$ and $g(x)$ is discontinuous at $x = 3$.

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Lyndt
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Lyndt

Updated on August 01, 2022

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  • Lyndt
    Lyndt over 1 year

    Find $f: \mathbb{R}\rightarrow \mathbb{R}$ and $g: \mathbb{R}\rightarrow \mathbb{R}$ such that $f$ is not continuous at 0, $g$ is not continuous at 3, and $f(x)g(x)$ is continuous everywhere.

    1. What does $f: \mathbb{R}\rightarrow \mathbb{R}$ mean? Does it necessarily mean that the domain of the function is $\mathbb{R}$?
    2. Where do I start to solve this problem? I was thinking $f(x)= \frac{x-3}{x}$ and $g(x)=\frac{x}{x-3}$ but is it correct if $f: \mathbb{R}\rightarrow \mathbb{R}$ and $g: \mathbb{R}\rightarrow \mathbb{R}$?
    • layman
      layman about 8 years
      Does $R$ represent the real numbers? If so, instead of R, try typing $\text{ \mathbb{R} }$. This will give you $\Bbb R$.
    • layman
      layman about 8 years
      Also, the notation $f : R \to R$ is a function (called $f$) that takes as input values from $R$, and spits out values in $R$. In general, $h : X \to Y$ is a function (called $h$) that takes as inputs values from the set $X$ and spits out as outputs values from the set $Y$.
    • layman
      layman about 8 years
      So, to solve this problem, you need to find a function $f: \Bbb R \to \Bbb R$ and a function $g : \Bbb R \to \Bbb R$ such that $f$ is not continuous at $0$, $g$ is not continuous at $3$, and $f(x)g(x)$ is continuous everywhere. The nice thing about functions from $\Bbb R$ to $\Bbb R$ is that we can visualize them. These are the functions you used in calculus! You can draw them in the $XY$-plane.
    • Martin Sleziak
      Martin Sleziak about 8 years
      Can you think at least of an example of a function $f$ from $\mathbb R$ to $\mathbb R$ which is not continuous at $0$?
    • Lyndt
      Lyndt about 8 years
      I need more details. I can't figure it out.
    • Martin Sleziak
      Martin Sleziak about 8 years
      The problem with $f(x)=\frac{x-3}x$ is the it is not defined for all real numbers. But for example $$f(x)= \begin{cases} \frac{x-3}x & \text{if }x\ne0, \\ 1 & \text{if }x=0. \end{cases} $$ would be a function from $\mathbb R$ to $\mathbb R$. Where is this function continuous and where not? Can you somehow use this example to get $f(x)$ and $g(x)$ as required?
    • Lyndt
      Lyndt about 8 years
      I can't figure out. Could you just give me the answer?
  • Axoren
    Axoren about 8 years
    My mistake, I misread the question. I was under the assumption that $f$ and $g$ were functions undefined at $0$ and $3$
  • layman
    layman about 8 years
    @Axoren Comment deleted. :)