Function's minimum on [1,5] is 0, achieved at x=1, found by analyzing critical points and endpoints.
Finding the Minimum Value of f(x)
To find the minimum value of f(x) = x² + x - 2 on the closed interval [1, 5], we can use two methods:
Method 1: Analyzing Critical Points and Endpoints
Find the critical points: A critical point is a point where the derivative f'(x) is equal to 0 or undefined.
For f(x) = x² + x - 2, the derivative is f'(x) = 2x + 1.
Therefore, f'(x) = 0 for x = -1/2, which is not within the interval [1, 5].
Evaluate f(x) at the endpoints and the critical point (if any):
f(1) = 1² + 1 - 2 = 0
f(5) = 5² + 5 - 2 = 28
Compare the function values: The minimum value of f(x) is the lowest among the values obtained in step 2. In this case, f(1) = 0 is the minimum value.
Method 2: Using the Extremum Value Theorem
The Extremum Value Theorem states that if a continuous function f(x) is defined on a closed interval [a, b], then it must have an absolute maximum and minimum within that interval.
Verify that f(x) is continuous on [1, 5]: f(x) is a polynomial function and is therefore continuous on all real numbers, including the closed interval [1, 5].
Apply the theorem: Since f(x) is continuous on [1, 5], it must have an absolute minimum within that interval. This minimum can occur either at a critical point or at an endpoint.
Follow the same steps as in method 1: Analyze the critical points (which are none in this case) and evaluate f(x) at the endpoints. The minimum value will be the lowest among the evaluated points.
Conclusion:
Both methods lead to the same conclusion: the minimum value of f(x) = x² + x - 2 on the closed interval [1, 5] is 0, which occurs at x = 1.
Visualization: The graph of f(x) = x² + x - 2 on the interval [1, 5] is shown below. The minimum point is clearly visible at x = 1.
Complete question:
What is the minimum value of the function f(x) = x² + x - 2 on the interval [1, 5]?