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The problem in the image is asking to find the equation of the tangent line to the graph of \( g(x) = f(2x^2 - 3) \) at the point where \( x = 2 \).

### Given Information:
- \( f(2) = 1 \), \( f'(2) = -6 \), \( f(5) = 1 \), and \( f'(5) = 5 \).
- \( g(x) = f(2x^2 - 3) \).

We are tasked with finding the equation of the tangent line to \( g(x) \) at \( x = 2 \).

### Steps:

1. **Find the point on the curve:**

   We know that the tangent line is calculated at \( x = 2 \). First, evaluate \( g(2) \):
   \[
   g(2) = f(2(2)^2 - 3) = f(8 - 3) = f(5)
   \]
   From the given information, \( f(5) = 1 \). So,
   \[
   g(2) = 1
   \]
   Hence, the point of tangency is \( (2, 1) \).

2. **Find the derivative of \( g(x) \):**

   To find the slope of the tangent line, we need \( g'(x) \), the derivative of \( g(x) = f(2x^2 - 3) \).
   
   Using the chain rule:
   \[
   g'(x) = f'(2x^2 - 3) \cdot \frac{d}{dx}(2x^2 - 3)
   \]
   The derivative of \( 2x^2 - 3 \) is \( 4x \). Therefore,
   \[
   g'(x) = f'(2x^2 - 3) \cdot 4x
   \]

   Now, substitute \( x = 2 \) into \( g'(x) \):
   \[
   g'(2) = f'(2(2)^2 - 3) \cdot 4(2) = f'(5) \cdot 8
   \]
   From the given information, \( f'(5) = 5 \). Thus,
   \[
   g'(2) = 5 \cdot 8 = 40
   \]

3. **Write the equation of the tangent line:**

   The tangent line has the form:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( (x_1, y_1) = (2, 1) \) and \( m = 40 \). Substituting these values:
   \[
   y - 1 = 40(x - 2)
   \]
   Simplifying:
   \[
   y = 40(x - 2) + 1 = 40x - 80 + 1 = 40x - 79
   \]

### Final Answer:
The equation of the tangent line is \( y = 40x - 79 \).

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

Here are 5 related questions to expand your understanding:
1. How does the chain rule apply to functions within functions, like in this problem?
2. What is the significance of the derivative in finding a tangent line?
3. How would the problem change if \( g(x) \) was a different composition of functions?
4. Can you find the tangent line to the graph of \( g(x) \) at \( x = 1 \)?
5. How does knowing the values of \( f'(x) \) help in solving the problem?

**Tip:** Always verify the function composition and derivative steps when using the chain rule, especially when evaluating complex expressions!

Let f be a differentiable function with f(2) = 1, f'(2) = -6, f(5) = 1, and f'(5) = 5. Let the function g(x) = f(2x^2 - 3). Write the equation of the line tangent to the graph of g at the point where x = 2.

Learn how to find the equation of the tangent line to the graph of g(x) = f(2x^2 - 3) at x = 2. This problem involves the application of the chain rule, differentiation, and tangent line formulas. Suitable for students in Grade 12 or early college.

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