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Let $y=f(x)$ be a function which is discontinuous for exactly $3$ values of $x$ but defined $\forall x~{\in}~\mathbb{R}$.
Let $y=g(x)$ is another differentiable function such that $y=f(x)g(x)$ is continuous $\forall x~{\in}~\mathbb{R}$.

Then find the minimum number of distinct real roots of the equation $g(x)g'(x)=0$

How to do it with proof? With mathematical intuition?
And also, how should i approach like this problems when faced? Thank you.

My work:

I just assumed the function to be $\frac{1}{(x-1)(x-2)(x-3)}$ and then g(x) trivially to deal easily $g(x)=(x-1)(x-2)(x-3)$. But I could not think it mathematically.

MarianD
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TerenceP
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1 Answers1

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The key is to make use of extended limit laws as discussed here. If $f$ is discontinuous at $a$ and $g$ and $fg$ are both continuous at $a$ then we must have $$\lim_{x\to a} g(x) =g(a) =0$$ Thus if $f$ is discontinuous at $a, b, c$ then $a, b, c$ are roots of $g$. Hence the desired minimum value is $3+2=5$ (two roots of $g'$ via Rolle).

Paramanand Singh
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