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We are tasked with finding \( f'(1) \) based on the given information. The function is defined as:

\[
f(x) = 3g(x) + h(x)
\]

We are also provided with these details:
- \( g(1) = -2 \) and \( g'(1) = 6 \)
- \( h(1) = 5 \) and \( h'(1) = 1 \)

### Step-by-Step Solution:

To find \( f'(x) \), we can differentiate \( f(x) = 3g(x) + h(x) \). Using the sum and constant multiple rules for derivatives, we get:

\[
f'(x) = 3g'(x) + h'(x)
\]

Now, substitute \( x = 1 \):

\[
f'(1) = 3g'(1) + h'(1)
\]

From the given information:
- \( g'(1) = 6 \)
- \( h'(1) = 1 \)

Substitute these values:

\[
f'(1) = 3(6) + 1 = 18 + 1 = 19
\]

Thus, \( f'(1) = 19 \).

Would you like more details or further clarification on this? 

Here are 5 related questions to deepen your understanding:
1. What rules of differentiation are used to compute \( f'(x) \)?
2. How would the result change if \( f(x) \) were a product of \( g(x) \) and \( h(x) \) instead of a sum?
3. How do the values of \( g(1) \) and \( h(1) \) affect the calculation of \( f'(1) \)?
4. Can you compute \( f'(x) \) for any general \( x \) given these functions?
5. What happens if one of the derivatives, like \( g'(1) \), were zero?

**Tip:** Always differentiate constants before applying any product or chain rules in composite functions!

Use the given information to find f'(1). f(x) = 3g(x) + h(x), g(1) = -2, g'(1) = 6, h(1) = 5, and h'(1) = 1.

Solve the differentiation problem f'(1) based on the given function f(x) = 3g(x) + h(x) and provided values for g(1), g'(1), h(1), and h'(1). This problem involves using the sum and constant multiple rules of differentiation. Suitable for Grades 10-12.

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