let f and g be functions that satisfy f'(2)=-2 and g'(2)=-3. find h'(2) for each function h given below: (E) h(x)=2f(x)+7g(x)-12. h'(2)=

To solve for \( h'(2) \) where \( h(x) = 2f(x) + 7g(x) - 12 \), we can use the following rules:

1. **Linearity of the derivative**: The derivative of a sum is the sum of the derivatives.
2. **Constant multiple rule**: The derivative of a constant times a function is the constant times the derivative of the function.

Thus, to differentiate \( h(x) = 2f(x) + 7g(x) - 12 \):

\[
h'(x) = 2f'(x) + 7g'(x) + 0
\]

Now, substitute \( x = 2 \):

\[
h'(2) = 2f'(2) + 7g'(2)
\]

We are given:

- \( f'(2) = -2 \)
- \( g'(2) = -3 \)

Substitute these values:

\[
h'(2) = 2(-2) + 7(-3)
\]
\[
h'(2) = -4 - 21
\]
\[
h'(2) = -25
\]

Thus, \( h'(2) = -25 \).

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Would you like further clarification or more details on this process? Here are 5 related questions for further exploration:

1. How do we find \( h'(x) \) for more complex functions involving \( f(x) \) and \( g(x) \)?
2. How does the chain rule apply if \( h(x) \) is a composition of functions?
3. What happens to the derivative if a function is multiplied by a variable instead of a constant?
4. Can you explain how product and quotient rules apply to functions like \( h(x) = f(x)g(x) \)?
5. How would you apply this to functions with higher derivatives?

**Tip:** Always remember that constants drop out when differentiating, which can simplify calculations significantly!

This calculus problem demonstrates how to find h'(2) for h(x) = 2f(x) + 7g(x) - 12 using key derivative rules such as the linearity of the derivative and the constant multiple rule. The solution shows how to apply these rules step by step. Suitable for college-level students.

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