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Find the critical points of the function f(x) = 3x^4 - 8x^3 + 6.
Find the critical points of the function f(x) = 3x^4 - 8x^3 + 6.
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Practice Questions
1 question
Q1
Find the critical points of the function f(x) = 3x^4 - 8x^3 + 6.
(0, 6)
(2, -2)
(1, 1)
(3, 0)
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f'(x) = 12x^3 - 24x^2. Setting f'(x) = 0 gives x^2(12x - 24) = 0, so x = 0 or x = 2. f(2) = 3(2^4) - 8(2^3) + 6 = -2.
Questions & Step-by-step Solutions
1 item
Q
Q: Find the critical points of the function f(x) = 3x^4 - 8x^3 + 6.
Solution:
f'(x) = 12x^3 - 24x^2. Setting f'(x) = 0 gives x^2(12x - 24) = 0, so x = 0 or x = 2. f(2) = 3(2^4) - 8(2^3) + 6 = -2.
Steps: 11
Show Steps
Step 1: Write down the function f(x) = 3x^4 - 8x^3 + 6.
Step 2: Find the derivative of the function, which is f'(x). The derivative tells us the slope of the function.
Step 3: Calculate the derivative: f'(x) = 12x^3 - 24x^2.
Step 4: Set the derivative equal to zero to find critical points: 12x^3 - 24x^2 = 0.
Step 5: Factor the equation: 12x^2(x - 2) = 0.
Step 6: Solve for x by setting each factor equal to zero: 12x^2 = 0 or x - 2 = 0.
Step 7: From 12x^2 = 0, we find x = 0. From x - 2 = 0, we find x = 2.
Step 8: The critical points are x = 0 and x = 2.
Step 9: To find the value of the function at x = 2, calculate f(2) = 3(2^4) - 8(2^3) + 6.
Step 10: Calculate f(2): f(2) = 3(16) - 8(8) + 6 = 48 - 64 + 6 = -10.
Step 11: The critical points are x = 0 and x = 2, with f(2) = -10.
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