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Show that 0 is a local minimum of f(x) = sin(x²).

User Ovicko
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Final answer:

To determine if 0 is a local minimum of f(x) = sin(x²), we find that the first derivative at x = 0 is 0, indicating a critical point. The second derivative is positive at this point, showing the function is concave up, thus confirming that x = 0 is a local minimum.

Step-by-step explanation:

To show that 0 is a local minimum for the function f(x) = sin(x²), we need to examine the behaviour of the function around x = 0. First, we calculate the first derivative f'(x) = 2x · cos(x²) to determine the critical points. Setting the first derivative to zero, we find that f'(0) = 0, indicating that x = 0 is a critical point.

Next, we evaluate the second derivative to determine the concavity around the critical point. The second derivative is f''(x) = -4x² sin(x²) + 2 cos(x²). At x = 0, this simplifies to f''(0) = 2, which is positive, showing that the function is concave up at x = 0. Therefore, x = 0 is a local minimum of the function.

User Amuliar
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