We are given the function $f(x) = (x^2 + 5x + 7)e^x$, and we are asked to find the $x$-coordinates where the tangent line to the graph of $f$ is horizontal.

To find where the tangent line is horizontal, we need to compute the derivative of $f(x)$, set it equal to zero, and solve for $x$.

Let's calculate the derivative $f'(x)$ using the product rule: $f(x) = (x^2 + 5x + 7) \cdot e^x$ Let $u(x) = x^2 + 5x + 7$ and $v(x) = e^x$. The product rule states: $f'(x) = u'(x) v(x) + u(x) v'(x)$ Where:

• $u'(x) = 2x + 5$
• $v'(x) = e^x$

Thus: $f'(x) = (2x + 5)e^x + (x^2 + 5x + 7)e^x$

We can factor out $e^x$: $f'(x) = e^x \left( (2x + 5) + (x^2 + 5x + 7) \right)$ Simplify the terms inside the parentheses: $f'(x) = e^x \left( x^2 + 7x + 12 \right)$

For the tangent line to be horizontal, we need $f'(x) = 0$. Since $e^x$ is never zero, we focus on solving: $x^2 + 7x + 12 = 0$

We can solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ For $x^2 + 7x + 12 = 0$, $a = 1$, $b = 7$, and $c = 12$. Plugging in these values: $x = \frac{-7 \pm \sqrt{7^2 - 4(1)(12)}}{2(1)}$ $x = \frac{-7 \pm \sqrt{49 - 48}}{2}$ $x = \frac{-7 \pm \sqrt{1}}{2}$ $x = \frac{-7 \pm 1}{2}$ So the two solutions are: $x = \frac{-7 + 1}{2} = -3 \quad \text{and} \quad x = \frac{-7 - 1}{2} = -4$

Thus, the $x$-coordinates where the tangent line is horizontal are $x = -3$ and $x = -4$.

Would you like more details on any step?

Here are 5 related questions:

1. How do we apply the product rule to more complex functions?
2. Why does factoring out $e^x$ simplify solving for where $f'(x) = 0$?
3. Can there be cases where no real solutions exist for horizontal tangents?
4. How does the quadratic formula work for equations with complex roots?
5. How would the result change if the exponential part were different, e.g., $e^{2x}$?

Tip: Always check if your function can be simplified before taking the derivative to make calculations easier.