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We are given that \( h(x) = f(g(x)) \), and the equations of the tangent lines at specific points for the functions \( f \) and \( g \). Let's solve the parts:

### Part a: Calculate \( h(4) \) and \( h'(4) \).

1. **Finding \( h(4) \):**

   Since \( h(x) = f(g(x)) \), to find \( h(4) \), we first need \( g(4) \). From the tangent line equation for \( g \) at the point \( (4,6) \), we see that \( g(4) = 6 \). Therefore:

   \[
   h(4) = f(g(4)) = f(6)
   \]

   From the tangent line equation for \( f \) at the point \( (6, 2) \), we know \( f(6) = 2 \). Hence:

   \[
   h(4) = 2
   \]

2. **Finding \( h'(4) \):**

   Using the chain rule, \( h'(x) = f'(g(x)) \cdot g'(x) \). So to find \( h'(4) \), we need both \( f'(g(4)) \) and \( g'(4) \).

   - From the tangent line of \( f \), the slope at \( (6, 2) \) is the derivative \( f'(6) \). The equation of the tangent line is \( y = -3x + 20 \), so the slope (which is \( f'(6) \)) is \( -3 \).
   
   - From the tangent line of \( g \), the slope at \( (4, 6) \) is the derivative \( g'(4) \). The equation of the tangent line is \( y = 4x - 10 \), so the slope (which is \( g'(4) \)) is \( 4 \).

   Now we can compute \( h'(4) \):

   \[
   h'(4) = f'(g(4)) \cdot g'(4) = f'(6) \cdot g'(4) = (-3) \cdot 4 = -12
   \]

Thus, \( h'(4) = -12 \).

---

### Part b: Determine the equation of the tangent line to the graph of \( h \) at the point where \( x = 4 \).

The equation of a tangent line is given by:

\[
y - y_1 = m(x - x_1)
\]

Here, \( x_1 = 4 \), \( y_1 = h(4) = 2 \), and the slope \( m = h'(4) = -12 \). Substituting these values into the equation:

\[
y - 2 = -12(x - 4)
\]

Simplifying:

\[
y - 2 = -12x + 48
\]

\[
y = -12x + 50
\]

So the equation of the tangent line is \( y = -12x + 50 \).

---

Let me know if you want further details or have any other questions!

### 5 Related Questions:
1. How do you apply the chain rule for composite functions like \( h(x) = f(g(x)) \)?
2. What is the significance of the derivative when determining the slope of a tangent line?
3. How do tangent line equations help in approximating function values near a point?
4. Can you always determine the derivative from a tangent line equation?
5. What other methods can be used to find \( h'(x) \) besides the chain rule?

### Tip:
When dealing with composite functions, always break the problem into smaller parts by focusing on the inner function first, then applying the chain rule to handle derivatives effectively.

Assume f and g are differentiable functions with h(x) = f(g(x)). Suppose the equation of the line tangent to the graph of g at the point (4,6) is y = 4x - 10 and the equation of the line tangent to the graph of f at (6,2) is y = -3x + 20. a. Calculate h(4) and h'(4). b. Determine an equation of the line tangent to the graph of h at the point on the graph where x = 4.

Learn how to calculate the tangent line for composite functions using differentiation and the chain rule. The problem involves differentiable functions f and g, with given tangent lines at specific points, and asks for the calculation of h(4), h'(4), and the equation of the tangent line at x = 4. Suitable for students in Grades 11-12 studying calculus.

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