Exercise · ZW-CONTINUOUS-FUNCTION-005 · ★★☆ Intermediate
Which value of k makes \(f(x)=\begin{cases}kx-3&\text{if }x\leq2\\ x^{2}-1&\text{if }x>2\end{cases}\) continuous at \(x = 2\)?
Choose one option.
Show answer
- A. \(3\)
Exercise · ZW-CONTINUOUS-FUNCTION-005 · ★★☆ Intermediate
Choose one option.
At x = 2 both expressions must agree: \(2k-3=2^{2}-1=3\). Then \(2k=6\) and k = 3.
A function \(f\) is continuous at \(a\) if \(f(a)\) is defined, its limit as \(x\) tends to \(a\) exists and equals \(f(a)\).