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$$ F(x) = \begin{cases} x + 1 & \text{for }x < 1 \\ 0 & \text{for }1 \leq x < 2 \\ 2-x & \text{for }x \geq 2 \end{cases} $$

Zubzub
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user428177
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    At $x=1$ there is a jump from $2$ to $0$, so it's discontinuous at $x=1$. At all other places the function is continuous. – rookie Mar 28 '17 at 09:03

2 Answers2

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We need to analyze two points :

  • When $x = 1$ : $\lim_{x\to 1^-} F(x) = 2$, $\ \lim_{x\to 1^+} F(x) = 0$
  • When $x = 2$ : $\lim_{x\to 2^-} F(x) = 0$, $\ \lim_{x\to 2^+} F(x) = 0$

Can you conclude ?

Zubzub
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In addition to the boundaries between the intervals $(x=1, x=2)$ where it has already been demonstrated that there is a discontinuity at $x=1$, according to your question you also need to consider all other $x$.

Since the function is comprised piecewise of 2 polynomials and a constant, which are continuous on their domains, it is true that $F(x)$ is continuous at all other values of $x$

PM.
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