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Determine the critical points of f(x) = x^4 - 4x^3 + 6.
Determine the critical points of f(x) = x^4 - 4x^3 + 6.
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Practice Questions
1 question
Q1
Determine the critical points of f(x) = x^4 - 4x^3 + 6.
x = 0, 3
x = 1, 2
x = 2, 3
x = 1, 3
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Setting f'(x) = 0 gives critical points at x = 1 and x = 2.
Questions & Step-by-step Solutions
1 item
Q
Q: Determine the critical points of f(x) = x^4 - 4x^3 + 6.
Solution:
Setting f'(x) = 0 gives critical points at x = 1 and x = 2.
Steps: 8
Show Steps
Step 1: Start with the function f(x) = x^4 - 4x^3 + 6.
Step 2: Find the derivative of the function, f'(x).
Step 3: Use the power rule to differentiate: f'(x) = 4x^3 - 12x^2.
Step 4: Set the derivative equal to zero: 4x^3 - 12x^2 = 0.
Step 5: Factor out the common term: 4x^2(x - 3) = 0.
Step 6: Set each factor equal to zero: 4x^2 = 0 and x - 3 = 0.
Step 7: Solve for x in each case: From 4x^2 = 0, we get x = 0; from x - 3 = 0, we get x = 3.
Step 8: The critical points are x = 0 and x = 3.
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