Question

1. (4pts) find the value of the constant ( k ) that makes ( f(x) ) continuous: ( f(x) = \begin{cases} x^2 - k & \text{if } x < 5 \\ kx + 2 & \text{if } x geq 5 end{cases} )

Explanation:

Step1: Set left/right limits equal

For continuity at $x=5$, $\lim_{x \to 5^-} f(x) = \lim_{x \to 5^+} f(x)$
Substitute the pieces: $5^2 - k = 5k + 2$

Step2: Simplify the equation

Calculate $5^2$ and rearrange terms:
$25 - k = 5k + 2$
$25 - 2 = 5k + k$
$23 = 6k$

Step3: Solve for k

Isolate $k$ by dividing both sides by 6:
$k = \frac{23}{6}$

Answer:

$\frac{23}{6}$