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!