If f(x)= ) x^2sin((1)/(x)),&ifxneq 0 6-2k,&ifx=0 is continuous at x=0 then the value of k=

To determine the value of that makes the function \( f(x) \) continuous at , we need to ensure that the limit of \( f(x) \) as approaches 0 is equal to \( f(0) \).First, let's find the limit of \( f(x) \) as approaches 0: Since \( \sin\left(\frac{1}{x}\right) \) oscillates between -1 and 1, we can use the Squeeze Theorem to evaluate this limit. We know that: Multiplying through by , we get: Taking the limit of each part as approaches 0, we get: Therefore, by the Squeeze Theorem: For \( f(x) \) to be continuous at , we must have: Given that \( f(0) = 6 - 2k \), we set this equal to the limit we found: Solving for : Thus, the value of that makes the function \( f(x) \) continuous at is .