Question

1. given that the function

$f(x)=\

$$\begin{cases}3x + 2, & x lt k \\\\ 2x - 3, & k \\leq x \\leq 8 \\end{cases}$$

$
is continuous on (-inf,8, find k.

Explanation:

Step1: Use the definition of continuity

For a function \( f(x) \) to be continuous at \( x = k \), \( \lim_{x
ightarrow k^{-}}f(x)=\lim_{x
ightarrow k^{+}}f(x)=f(k) \).
Since \( \lim_{x
ightarrow k^{-}}f(x)=3k + 2 \) (using \( y=3x + 2 \) for \( xightarrow k^{+}}f(x)=2k-3 \) (using \( y = 2x-3 \) for \( k\leq x\leq8 \)) and \( f(k)=2k - 3 \).
Set \( 3k+2=2k - 3 \).

Step2: Solve the equation for \( k \)

Subtract \( 2k \) from both sides of the equation \( 3k+2=2k - 3 \).
We get \( 3k-2k+2=2k-2k - 3 \), which simplifies to \( k+2=-3 \).
Then subtract 2 from both sides: \( k=-3 - 2 \).

Answer:

\( k=-5 \)