find k so that the following function is continuous.\nf(x)=\begin{cases}kx&\text{if }0leq x <…

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

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

Answer

Explanation:

Step1: Recall continuity condition

For a function to be continuous at (x = 1), (\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{+}}f(x)).

Step2: Calculate left - hand limit

(\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{-}}kx). Substituting (x = 1) into (kx), we get (k\times1=k).

Step3: Calculate right - hand limit

(\lim_{x\rightarrow1^{+}}f(x)=\lim_{x\rightarrow1^{+}}2x^{2}). Substituting (x = 1) into (2x^{2}), we have (2\times1^{2}=2).

Step4: Equate the limits

Since (\lim_{x\rightarrow1^{-}}f(x)=\lim_{x\rightarrow1^{+}}f(x)) for continuity, then (k = 2).

Answer:

(2)