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For which value of a is the function f(x) = x^3 - 3ax^2 + 3a^2x + 1 differentiab
For which value of a is the function f(x) = x^3 - 3ax^2 + 3a^2x + 1 differentiable at x = 1?
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For which value of a is the function f(x) = x^3 - 3ax^2 + 3a^2x + 1 differentiable at x = 1?
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Setting f'(1) = 0 gives a = 1 for differentiability at x = 1.
Questions & Step-by-step Solutions
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Q
Q: For which value of a is the function f(x) = x^3 - 3ax^2 + 3a^2x + 1 differentiable at x = 1?
Solution:
Setting f'(1) = 0 gives a = 1 for differentiability at x = 1.
Steps: 5
Show Steps
Step 1: Write down the function f(x) = x^3 - 3ax^2 + 3a^2x + 1.
Step 2: Find the derivative of the function, f'(x).
Step 3: Substitute x = 1 into the derivative to find f'(1).
Step 4: Set f'(1) equal to 0 to find the value of a that makes the function differentiable at x = 1.
Step 5: Solve the equation from Step 4 to find a.
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