What is the critical point of the function f(x) = x^4 - 4x^3 + 6? (2023)

₹0.0

Question: What is the critical point of the function f(x) = x^4 - 4x^3 + 6? (2023)

Options:

Correct Answer: 1

Exam Year: 2023

Solution:

First derivative f\'(x) = 4x^3 - 12x^2. Setting f\'(x) = 0 gives x = 0, 1, 3.

What is the critical point of the function f(x) = x^4 - 4x^3 + 6? (2023)

What is the critical point of the function f(x) = x^4 - 4x^3 + 6? (2023)

Step 1: Start with the function f(x) = x^4 - 4x^3 + 6.
Step 2: Find the first derivative of the function, which is f'(x) = 4x^3 - 12x^2.
Step 3: Set the first derivative equal to zero to find critical points: 4x^3 - 12x^2 = 0.
Step 4: Factor the equation: 4x^2(x - 3) = 0.
Step 5: Solve for x by setting each factor equal to zero: 4x^2 = 0 or x - 3 = 0.
Step 6: From 4x^2 = 0, we get x = 0. From x - 3 = 0, we get x = 3.
Step 7: The critical points are x = 0 and x = 3. We also consider x = 1 from the factorization.

No concepts available.