Find f′(−8)f′(−8), given thatf(x)=5x⋅h(x)h(−8)=−6h′(−8)=3f(x)=5x⋅h(x)h(−8)=−6h′(−8)=3

−90−90

−149−149

−150−150

−160−160

90

We are given the function \( f(x) = 5x \cdot h(x) \), and the values \( h(-8) = -6 \) and \( h'(-8) = 3 \). We are tasked with finding \( f'(-8) \).

### Step 1: Apply the Product Rule
The product rule for differentiation states that:
\[
f'(x) = \frac{d}{dx} [5x \cdot h(x)] = 5 \cdot h(x) + 5x \cdot h'(x)
\]
We need to find \( f'(-8) \), so we will substitute \( x = -8 \) into this formula.

### Step 2: Substitute values
From the problem, we know:
- \( h(-8) = -6 \)
- \( h'(-8) = 3 \)

Using these in the expression for \( f'(x) \):
\[
f'(-8) = 5 \cdot h(-8) + 5 \cdot (-8) \cdot h'(-8)
\]
Substitute the known values:
\[
f'(-8) = 5 \cdot (-6) + 5 \cdot (-8) \cdot 3
\]
\[
f'(-8) = -30 + 5 \cdot (-8) \cdot 3
\]
\[
f'(-8) = -30 + (-120)
\]
\[
f'(-8) = -150
\]

### Final Answer:
\[
f'(-8) = -150
\]

Thus, the correct answer is \( \boxed{-150} \).

Would you like further details or have any other questions?

Here are five related questions you can explore:
1. How do you differentiate a product of two functions?
2. How does the product rule apply if both functions are complicated expressions?
3. What are the steps to apply the chain rule in derivatives?
4. What happens if one function is constant during differentiation?
5. How does the value of \( h'(x) \) affect the slope of \( f(x) \)?

**Tip:** The product rule is a handy tool when differentiating functions that are products of two or more terms. Always differentiate each term independently and then combine the results.

Find f′(−8), given that f(x)=5x⋅h(x), h(−8)=−6, h′(−8)=3

Find f'(-8) for the function f(x) = 5x ⋅ h(x) using the product rule of differentiation. Given h(-8) = -6 and h'(-8) = 3, this problem demonstrates key calculus concepts suitable for high school students in Grades 11-12.

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