Question

find k so that the following function is continuous.
f(x)=\begin{cases}kx&\text{if }0leq x < 1\\2x^{2}&\text{if }1leq xend{cases}
enter dne if no such k - value is possible.
k =

Explanation:

Step1: Recall continuity condition

For a function to be continuous at \(x = 1\), \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)\).

Step2: Calculate left - hand limit

\(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{-}}kx\). Substituting \(x = 1\) into \(kx\), we get \(k\times1=k\).

Step3: Calculate right - hand limit

\(\lim_{x
ightarrow1^{+}}f(x)=\lim_{x
ightarrow1^{+}}2x^{2}\). Substituting \(x = 1\) into \(2x^{2}\), we have \(2\times1^{2}=2\).

Step4: Equate the limits

Since \(\lim_{x
ightarrow1^{-}}f(x)=\lim_{x
ightarrow1^{+}}f(x)\) for continuity, then \(k = 2\).

Answer:

\(2\)