To solve for $h'(2)$, where $h(x) = \frac{f(x) + 1}{g(x) + 3x}$, we need to apply the Quotient Rule. The Quotient Rule states that if $h(x) = \frac{u(x)}{v(x)}$, then $h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.

Step 1: Identify $u(x)$ and $v(x)$

For $h(x) = \frac{f(x) + 1}{g(x) + 3x}$, we define:

• $u(x) = f(x) + 1$
• $v(x) = g(x) + 3x$

Then,

• $u'(x) = f'(x)$ (since the derivative of 1 is 0),
• $v'(x) = g'(x) + 3$ (since the derivative of $3x$ is 3).

Step 2: Determine $f(2)$, $f'(2)$, $g(2)$, and $g'(2)$ from the graphs

Using the graphs:

• For $f(x)$ at $x = 2$: The point on $f$ at $x = 2$ is $(2, 3)$, so $f(2) = 3$.

• For $f'(2)$: The slope between $x = 1$ and $x = 2$ for $f(x)$ is $\frac{3 - 2}{2 - 1} = 1$, so $f'(2) = 1$.

• For $g(x)$ at $x = 2$: The point on $g$ at $x = 2$ is $(2, 5)$, so $g(2) = 5$.

• For $g'(2)$: The slope between $x = 1$ and $x = 2$ for $g(x)$ is $\frac{5 - 6}{2 - 1} = -1$, so $g'(2) = -1$.

Step 3: Substitute into the Quotient Rule formula

Now, calculate $h'(2)$ using: $h'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{(v(x))^2}$

Substituting $x = 2$:

• $u(2) = f(2) + 1 = 3 + 1 = 4$
• $u'(2) = f'(2) = 1$
• $v(2) = g(2) + 3 \cdot 2 = 5 + 6 = 11$
• $v'(2) = g'(2) + 3 = -1 + 3 = 2$

Thus, $h'(2) = \frac{(1)(11) - (4)(2)}{(11)^2} = \frac{11 - 8}{121} = \frac{3}{121}$

Conclusion

The value of $h'(2)$ is $\frac{3}{121}$, which matches none of the answer choices exactly. However, the closest answer would be:

• $\frac{9}{100}$ (Option B), approximating the decimal equivalence.

Please let me know if you'd like further clarification!

Questions for further exploration:

1. What is the Quotient Rule, and how does it apply to different functions?
2. How do you find the derivative of piecewise functions?
3. How do slopes between points on graphs relate to derivatives?
4. Can a derivative be undefined based on graph shape?
5. How do approximations affect interpretations in multiple-choice tests?

Tip: Always double-check points on a graph when interpreting slopes, as small errors can impact derivative accuracy.