Final answer:
The function f(x) = x^2 - 2x + 3 has a single critical point at x = 1, determined by setting its derivative (2x - 2) to zero. The behavior analysis using the first derivative test shows that this critical point is a minimum because the function decreases before x = 1 and increases after. The second derivative test confirms that the function is concave up, further proving that x = 1 is a minimum.
Step-by-step explanation:
To find the critical points of the function f(x) = x^2 - 2x + 3, we need to determine where the derivative of the function is either zero or undefined. The derivative of f(x) is f'(x) = 2x - 2. Critical points occur when f'(x) = 0, so we set the derivative equal to zero and solve for x:
To analyze the behavior of f(x) at this critical point, we can use the first derivative test. We check the sign of f'(x) on either side of the critical point:
- For values of x less than 1, f'(x) is negative, which means f(x) is decreasing.
- For values of x greater than 1, f'(x) is positive, which indicates f(x) is increasing.
This suggests that the critical point at x = 1 is a minimum for f(x). To further confirm, we can use the second derivative, f''(x) = 2, which is positive for all x, indicating that the function is concave up and thus the critical point is indeed a minimum.
The function has a single critical point at x = 1, and based on our analysis, this point represents a minimum point on the graph of the function.