A Step-by-Step Guide to Finding the Absolute Minimum in Calculus: Illustrated Example

absolute minimum

The concept of absolute minimum is often discussed in calculus and refers to the lowest point on the graph of a function over a specific interval

The concept of absolute minimum is often discussed in calculus and refers to the lowest point on the graph of a function over a specific interval.

To find the absolute minimum, we need to follow these steps:

1. Determine the domain: Find the interval over which you are looking for the absolute minimum. This could be specified in the problem or given by a specific range or function.

2. Find the critical points: These are the points where the derivative of the function is either zero or undefined. To find them, take the derivative of the function and set it equal to zero, then solve for the variable. If there is a point where the denominator of the derivative becomes zero, this point will also be considered a critical point.

3. Evaluate the function at the critical points: Substitute each critical point into the original function to find the corresponding y-values.

4. Evaluate the function at the endpoints of the interval: If the interval is closed (including its endpoints), evaluate the function at both endpoints.

5. Compare the values obtained: Compare all the values obtained in steps 3 and 4. The smallest y-value among these points is the absolute minimum.

It’s important to note that the absolute minimum might not exist if the function is unbounded below or if there is no interval specified.

Let’s illustrate this process with an example:

Find the absolute minimum of the function f(x) = x^3 – 6x^2 + 9x + 2 over the closed interval [0, 5].

Step 1: The domain is the closed interval [0, 5].

Step 2: Take the derivative of f(x) and set it equal to zero:
f'(x) = 3x^2 – 12x + 9 = 0

Solving this quadratic equation, we get x = 1 and x = 3 as critical points.

Step 3: Evaluate the function at the critical points:
f(1) = 1^3 – 6(1)^2 + 9(1) + 2 = 6
f(3) = 3^3 – 6(3)^2 + 9(3) + 2 = 2

Step 4: Evaluate the function at the endpoints:
f(0) = 0^3 – 6(0)^2 + 9(0) + 2 = 2
f(5) = 5^3 – 6(5)^2 + 9(5) + 2 = -18

Step 5: Compare the values:
The function has an absolute minimum at x = 5, with a y-value of -18.

Therefore, the absolute minimum of the function f(x) = x^3 – 6x^2 + 9x + 2 over the interval [0, 5] is -18 at x = 5.

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