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Find the critical points of f(x) = x^4 - 8x^2 + 16. (2021)
Find the critical points of f(x) = x^4 - 8x^2 + 16. (2021)
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Practice Questions
1 question
Q1
Find the critical points of f(x) = x^4 - 8x^2 + 16. (2021)
(0, 16)
(2, 0)
(4, 0)
(1, 15)
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f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 0 is a critical point.
Questions & Step-by-step Solutions
1 item
Q
Q: Find the critical points of f(x) = x^4 - 8x^2 + 16. (2021)
Solution:
f'(x) = 4x^3 - 16x. Setting f'(x) = 0 gives x = 0, ±2. f(2) = 0 is a critical point.
Steps: 7
Show Steps
Step 1: Write down the function f(x) = x^4 - 8x^2 + 16.
Step 2: Find the derivative of the function, f'(x). The derivative is f'(x) = 4x^3 - 16x.
Step 3: Set the derivative equal to zero to find critical points: 4x^3 - 16x = 0.
Step 4: Factor the equation: 4x(x^2 - 4) = 0.
Step 5: Solve for x by setting each factor to zero: 4x = 0 gives x = 0, and x^2 - 4 = 0 gives x = ±2.
Step 6: List the critical points found: x = 0, x = 2, and x = -2.
Step 7: To confirm, evaluate f(2) = 0, which is a critical point.
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