Question

find \\(k\\) so that the following function is continuous:

\\f(x) = \

$$\begin{cases} kx & \\text{if } 0 \\le x < 5 \\\\ 2x^2 & \\text{if } 5 \\le x \\end{cases}$$

\\

\\(k =\\)

Explanation:

Identify the condition for continuity

To ensure the piecewise function \(f(x)\) is continuous everywhere, we must analyze the boundary point where the definition changes, which is at \(x = 5\). For a function to be continuous at a point, the limit from the left, the limit from the right, and the function value must all be equal. We apply the Continuity of Functions concept.

Evaluate the left-hand limit

We find the limit of \(f(x)\) as \(x\) approaches \(5\) from the left using the first piece of the function:

$$ \lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} (kx) = 5k $$

Evaluate the right-hand limit and function value

We find the limit of \(f(x)\) as \(x\) approaches \(5\) from the right and the value of the function at \(x = 5\) using the second piece:

$$ \lim_{x \to 5^+} f(x) = f(5) = 2(5)^2 = 2(25) = 50 $$

Solve for the constant k

For \(f(x)\) to be continuous at \(x = 5\), the left-hand limit must equal the right-hand limit:

$$ 5k = 50 $$

Solving for \(k\):

$$ k = 10 $$

Answer:

Find \(k\) so that the following function is continuous:

$$ f(x) =
LATEXBLOCK0
$$

\(k =\) <blank>10</blank>