For which value of a is the function f(x) = x^3 - 3ax + 2 differentiable at x =

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Question: For which value of a is the function f(x) = x^3 - 3ax + 2 differentiable at x = 1?

Options:

Correct Answer: 1

Solution:

Setting f\'(1) = 0 gives a = 1, ensuring differentiability at that point.

For which value of a is the function f(x) = x^3 - 3ax + 2 differentiable at x = 1?

For which value of a is the function f(x) = x^3 - 3ax + 2 differentiable at x = 1?

Step 1: Understand that a function is differentiable at a point if its derivative exists at that point.
Step 2: Find the derivative of the function f(x) = x^3 - 3ax + 2.
Step 3: The derivative f'(x) is calculated as f'(x) = 3x^2 - 3a.
Step 4: Substitute x = 1 into the derivative to find f'(1). This gives f'(1) = 3(1)^2 - 3a = 3 - 3a.
Step 5: For the function to be differentiable at x = 1, we need f'(1) to equal 0. Set the equation 3 - 3a = 0.
Step 6: Solve the equation 3 - 3a = 0 for a. This gives 3a = 3, so a = 1.
Step 7: Conclude that the function f(x) is differentiable at x = 1 when a = 1.

Differentiability – The function must have a defined derivative at the point of interest, which in this case is x = 1.
Finding Derivatives – Calculating the derivative of the function and evaluating it at a specific point.
Setting Derivative to Zero – To find critical points or ensure a function is flat at a point, setting the derivative equal to zero is necessary.