Questions: A function f has derivative f'(x)=x(x+3) e^x, and a critical point at x=0. Classify the critical point. LOCAL MAXIMUM LOCAL MINIMUM NOT A LOCAL EXTREMUM

Solution Steps

To classify the critical point at \( x = 0 \), we need to analyze the second derivative \( f''(x) \). If \( f''(0) > 0 \), the critical point is a local minimum. If \( f''(0) < 0 \), the critical point is a local maximum. If \( f''(0) = 0 \), the test is inconclusive.

The first derivative of the function is given by

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

To classify the critical point at \( x = 0 \), we need to compute the second derivative \( f''(x) \). The second derivative is calculated as follows:

\[ f''(x) = f'(x) + \text{additional terms from the product rule}. \]

After simplification, we find that

\[ f''(0) = 3. \]

To classify the critical point at \( x = 0 \), we evaluate \( f''(0) \):

• If \( f''(0) > 0 \), then \( x = 0 \) is a local minimum.
• If \( f''(0) < 0 \), then \( x = 0 \) is a local maximum.
• If \( f''(0) = 0 \), the test is inconclusive.

Since \( f''(0) = 3 > 0 \), we conclude that \( x = 0 \) is a local minimum.

Final Answer