Exercise · ZW-CONTINUOUS-FUNCTION-004 · ★★☆ Intermediate
Which value of k makes \(f(x)=\begin{cases}x+1&\text{if }x<1\\ kx^{2}&\text{if }x\geq1\end{cases}\) continuous at \(x = 1\)?
Choose one option.
Show answer
- B. \(2\)
Explanation
The one-sided limits at 1 must agree with f(1). From the left: \(1+1=2\); from the right and at 1: \(k\cdot1^{2}=k\). So k = 2.
Review the concept: Continuous function
A function \(f\) is continuous at \(a\) if \(f(a)\) is defined, its limit as \(x\) tends to \(a\) exists and equals \(f(a)\).
\[\lim_{x\to a}f(x)=f(a)\]
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