We are given the graphs of two functions $f(x)$ and $g(x)$, and the function $h(x) = \frac{f(x) + 4}{g(x) + 2x}$. We are asked to find $h'(3)$, the derivative of $h(x)$ at $x = 3$.

Step 1: Differentiate $h(x)$

We will use the quotient rule to differentiate $h(x)$. The quotient rule states that for a function of the form $\frac{u(x)}{v(x)}$, the derivative is given by: $h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ In this case, $u(x) = f(x) + 4$ and $v(x) = g(x) + 2x$.

Now, let's find the derivatives:

• $u'(x) = f'(x)$ because the derivative of the constant 4 is 0.
• $v'(x) = g'(x) + 2$ because the derivative of $2x$ is 2.

Thus, applying the quotient rule: $h'(x) = \frac{(f'(x))(g(x) + 2x) - (f(x) + 4)(g'(x) + 2)}{(g(x) + 2x)^2}$

Step 2: Calculate values at $x = 3$

We need to find the values of $f(3)$, $f'(3)$, $g(3)$, and $g'(3)$ using the graphs.

• From the graph of $f(x)$, we can see that $f(3) = 5$.

• The slope of $f(x)$ from $x = 2$ to $x = 4$ is: $f'(3) = \frac{f(4) - f(2)}{4 - 2} = \frac{3 - 4}{2} = \frac{-1}{2}$

• From the graph of $g(x)$, we observe that $g(3) = 5$.

• The slope of $g(x)$ from $x = 2$ to $x = 4$ is: $g'(3) = \frac{g(4) - g(2)}{4 - 2} = \frac{1 - 5}{2} = \frac{-4}{2} = -2$

Step 3: Plug values into the derivative formula

Now substitute $x = 3$, along with the values we calculated, into the formula for $h'(x)$: $h'(3) = \frac{\left( \frac{-1}{2} \right)(5 + 6) - (5 + 4)(-2 + 2)}{(5 + 6)^2}$ Simplifying the terms: $h'(3) = \frac{\left( \frac{-1}{2} \right)(11) - (9)(0)}{11^2} = \frac{\frac{-11}{2}}{121} = \frac{-11}{2 \times 121} = \frac{-11}{242}$

Thus: $h'(3) = \frac{-1}{22}$

This does not match the available answer choices. Upon review, let's check if there was an arithmetic mistake.

The correct differentiation