How does f(x) = x + 3 if x < 1; 2x - 1 if x ≥ 1 behave at x = 1?

Continuity & Differentiability · Question 16 of 125

Question 16

MediumOriginal question

How does $f(x) = \begin{cases} x + 3, & x < 1 \\ 2x - 1, & x \ge 1 \end{cases}$ behave at $x = 1$ ?

Continuous at the pointCorrect answerYour answerNot continuous: the limit exists but differs from the valueCorrect answerYour answerNot continuous: the function has no value thereCorrect answerYour answerNot continuous: the left and right limits differCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer D

Correct answer: D

Continuity at $a$ needs the left limit, the right limit and $f(a)$ to be equal.
Here the left limit is $4$ , the right limit $1$ and $f(1) = 1$ .

Worth noting

The left piece tends to 4 and the right gives 1: they do not meet.

Concept: Testing continuity at a point
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsContinuity at a point

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How does f(x)={x+3,x<12x−1,x≥1f(x) = \begin{cases} x + 3, & x < 1 \\ 2x - 1, & x \ge 1 \end{cases}f(x)={x+3,2x−1,​x<1x≥1​ behave at x=1x = 1x=1?

Explanation

How does $f(x) = \begin{cases} x + 3, & x < 1 \\ 2x - 1, & x \ge 1 \end{cases}$ behave at $x = 1$ ?