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$$f(x)=\lvert x^2 + 2x - 3\rvert$$

In the above function we can show continuity at a point by finding the left hand and right hand limits at that point! But how do we show that, this function is continuous at everywhere?

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  • $\begingroup$ Polynomials are always continuous everywhere. A composition of two continuous functions leads to a continuous function. $\endgroup$ – Andrew Li Apr 1 '18 at 0:45
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The composition of two functions that are continuous everywhere is continuous everywhere. Since $|\cdot|$ and $x^2+2x−3$ are continuous everywhere, and your $f(x)$ is the composition of $|\cdot|$ and $x^2+2x−3$, you can conclude.

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  • $\begingroup$ @TobyMak - It is a continuous function for any $x \in \mathbb{R}$. $\endgroup$ – Taroccoesbrocco Apr 1 '18 at 0:27
  • $\begingroup$ Sorry, I confused being 'continuous' with being 'differentiable'. $\endgroup$ – Toby Mak Apr 1 '18 at 0:28
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Note that $$f(x)=\lvert x^2 + 2x - 3\rvert$$

is the composite function of the absolute value function and the polynomial function $$x^2 + 2x - 3$$ which are continuous everywhere.

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By definition, a function is continuous "everywhere" (on its domain), if it is continuous at each point of the domain.

So, as you can show continuity of $f(x)=\lvert x^2 + 2x - 3\rvert$ at each point of its domain (using limits for example), you can conclude that your function is continuous "everywhere".

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