We are asked to find the derivative of the function $f(x) = \frac{g(x)}{h(x)}$ at $x = 3$, denoted as $f'(3)$.

Step 1: Apply the Quotient Rule

The derivative of a quotient of two functions, $f(x) = \frac{g(x)}{h(x)}$, is given by the quotient rule:

$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$

We will apply this rule to find $f'(3)$.

Step 2: Find $g(3), g'(3), h(3),$ and $h'(3)$ from the graphs.

1. For $g(x)$:

   • From the graph, at $x = 3$, we can see that $g(3) = 3$.
   • The slope of the line $g(x)$ is constant (it's a straight line), so we can calculate the slope between two points, say $(1, 7)$ and $(4, 2)$. The slope is:

   $g'(x) = \frac{7 - 2}{1 - 4} = \frac{5}{-3} = -\frac{5}{3}$

   Hence, $g'(3) = -\frac{5}{3}$.

2. For $h(x)$:

   • From the graph, at $x = 3$, we see that $h(3) = 5$.
   • To find $h'(3)$, notice that $h(x)$ has a vertex at $x = 3$, which means the slope $h'(3) = 0$ because the graph is flat at this point.

Step 3: Apply the values to the Quotient Rule

Now that we have all the values:

• $g(3) = 3$
• $g'(3) = -\frac{5}{3}$
• $h(3) = 5$
• $h'(3) = 0$

Substitute these into the quotient rule:

$f'(3) = \frac{\left(-\frac{5}{3}\right)(5) - (3)(0)}{(5)^2} = \frac{-\frac{25}{3}}{25} = -\frac{1}{3}$

Final Answer:

$f'(3) = -\frac{1}{3}$

Do you want any further clarifications or details? Here are 5 related questions you might want to explore:

1. What is the general form of the quotient rule for derivatives?
2. How do you find the slope of a line from a graph?
3. What happens when the derivative at a point is zero?
4. Can the quotient rule be applied to functions that are not linear?
5. What are the limitations of the quotient rule when the denominator is zero?

Tip: When applying the quotient rule, ensure that both the numerator and denominator are differentiable at the point of interest to avoid complications.