Find all the relative extrema of f(x) = x = x3 – 3x2 + 4.
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To find the relative extrema of a function, we need to find the points where the derivative of the function is zero or undefined. Let's start by finding the derivative of the given function.
Given function: f(x) = x^3 – 3x^2 + 4
Step 1: Find the derivative of the function f(x).
f'(x) = d/dx [x^3 – 3x^2 + 4]
= 3x^2 - 6x
Step 2: Set the derivative equal to zero and solve for x.
3x^2 - 6x = 0
Step 3: Factor out x from the equation.
3x(x - 2) = 0
Step 4: Set each factor equal to zero and solve for x.
3x = 0 or x - 2 = 0
For the first factor:
3x = 0
x = 0
For the second factor:
x - 2 = 0
x = 2
Therefore, we have found the critical points where the derivative is zero: x = 0 and x = 2.
Step 5: Determine the nature of extrema at the critical points.
To determine whether these points are relative extrema, we need to examine the sign changes in the derivative around these points.
Step 6: Create a sign chart by choosing test points for each interval.
Let's choose the test points: x = -1, x = 1, and x = 3.
Step 7: Evaluate the derivative at each test point and determine the sign.
For x = -1:
f'(-1) = 3(-1)^2 - 6(-1) = 3 + 6 = 9 (positive)
For x = 1:
f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3 (negative)
For x = 3:
f'(3) = 3(3)^2 - 6(3) = 27 - 18 = 9 (positive)
Step 8: Analyze the sign chart to determine the nature of extrema.
From the sign chart, we can observe that the sign of the derivative changes from positive to negative at x = 2. This indicates a relative maximum.
Similarly, the sign of the derivative changes from negative to positive at x = 0. This indicates a relative minimum.
Therefore, the relative extrema for the function f(x) = x^3 – 3x^2 + 4 are:
Relative maximum at x = 2,
Relative minimum at x = 0.