Question:

Step 1: Interpret the given limit condition to find f'(0). We are given that $\lim_{x \to 0} \frac{f(x) - 5}{x} = 4$. Since $f(0) = 5$, we can recognize this limit as the definition of the derivative of $f$ at $x = 0$: $$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x} = 4$$ Step 2: Apply the product rule to find g'(x). Given that $g(x) = (x^2 + 2x + 3)f(x)$, we use the product rule: $(uv)' = u'v + uv'$. Let $u(x) = x^2 + 2x + 3$ and $v(x) = f(x)$. Then: $$u'(x) = 2x + 2$$ $$v'(x) = f'(x)$$ Therefore: $$g'(x) = (2x + 2)f(x) + (x^2 + 2x + 3)f'(x)$$ Step 3: Evaluate g'(x) at x = 0. Substituting $x = 0$ into the derivative: $$g'(0) = (2(0) + 2)f(0) + (0^2 + 2(0) + 3)f'(0)$$ $$g'(0) = (2)f(0) + (3)f'(0)$$ Step 4: Substitute the known values and calculate. We know that $f(0) = 5$ and $f'(0) = 4$. Substituting these values: $$g'(0) = 2(5) + 3(4)$$ $$g'(0) = 10 + 12$$ $$g'(0) = 22$$ **Final Answer:** $g'(0) = 22$, which corresponds to **Option 1: 22**.