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To solve question 8, we need to find the value of \( k \) that makes \( p(x) \) continuous at \( x = 3 \).

The piecewise function is given as:

\[
p(x) = \begin{cases} 
3 - 2g(x), & x < 3 \\
5k + f(x), & x \geq 3 
\end{cases}
\]

To ensure continuity at \( x = 3 \), the left-hand limit (as \( x \to 3^- \)) must equal the right-hand limit (as \( x \to 3^+ \)), and both must equal the value of the function at \( x = 3 \).

### Step 1: Evaluate the left-hand limit as \( x \to 3^- \)
For \( x < 3 \), the function is \( p(x) = 3 - 2g(x) \). Therefore:
\[
\lim_{x \to 3^-} p(x) = 3 - 2g(3)
\]

From the graph of \( g(x)

Solve question 8: Let p be the piecewise defined function p(x) = 3 - 2g(x) for x < 3 and p(x) = 5k + f(x) for x ≥ 3. Find the value of k that makes p continuous at x = 3.

Learn how to solve for the value of k that ensures the continuity of the piecewise function p(x) = 3 - 2g(x) for x < 3 and p(x) = 5k + f(x) for x ≥ 3. This problem uses key concepts of limits and continuity, and is suitable for students in Grades 10-12.

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