How does f(x) = 2x + 1 if x ≤ 1; x^2 + 2 if x > 1 behave at x = 1?

Question 1

EasyOriginal question

How does $f(x) = \begin{cases} 2x + 1, & x \le 1 \\ x^{2} + 2, & x > 1 \end{cases}$ behave at $x = 1$ ?

Not continuous: the left and right limits differCorrect answerYour answerContinuous at the pointCorrect answerYour answerNot continuous: the limit exists but differs from the valueCorrect answerYour answerNot continuous: the function has no value thereCorrect answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer B

Correct answer: B

Continuity at $a$ needs the left limit, the right limit and $f(a)$ to be equal.
Here the left limit is $3$ , the right limit $3$ and $f(1) = 3$ .

Worth noting

Both pieces give 3 at $x = 1$ ; a change of formula is not a break if the pieces meet.

Concept: Testing continuity at a point
Typical timeTypical: 60 s
Your timeYours: —

The rule this testsContinuity at a point

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How does f(x)={2x+1,x≤1x2+2,x>1f(x) = \begin{cases} 2x + 1, & x \le 1 \\ x^{2} + 2, & x > 1 \end{cases}f(x)={2x+1,x2+2,​x≤1x>1​ behave at x=1x = 1x=1?

Explanation

How does $f(x) = \begin{cases} 2x + 1, & x \le 1 \\ x^{2} + 2, & x > 1 \end{cases}$ behave at $x = 1$ ?