Answer is: option1

Solution:

We are given:

\[ f'(x) = (x^3 + 2)e^x \]

We are to find the x-coordinate of the inflection point of the graph of \( f \), which means we need to find where the second derivative changes concavity — i.e., where \( f''(x) = 0 \) and changes sign.

We use the product rule:

\[ f''(x) = \frac{d}{dx} \left[ (x^3 + 2)e^x \right] \]

\[ f''(x) = \frac{d}{dx}(x^3 + 2) \cdot e^x + (x^3 + 2) \cdot \frac{d}{dx}(e^x) \]

\[ f''(x) = (3x^2)e^x + (x^3 + 2)e^x \]

Factor out \( e^x \):

\[ f''(x) = e^x(3x^2 + x^3 + 2) = e^x(x^3 + 3x^2 + 2) \]

Set \( f''(x) = 0 \) and solve

\[ e^x(x^3 + 3x^2 + 2) = 0 \]

\[ x^3 + 3x^2 + 2 = 0 \]

Using a graphing calculator or desmos we get x ≈ -3.19582