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Find the value of k for which the function f(x) = kx^2 + 2x + 1 is differentiabl
Find the value of k for which the function f(x) = kx^2 + 2x + 1 is differentiable at x = 0.
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Q1
Find the value of k for which the function f(x) = kx^2 + 2x + 1 is differentiable at x = 0.
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f'(x) = 2kx + 2. At x = 0, f'(0) = 2. The function is differentiable for any k, but k = 0 gives a constant function.
Questions & Step-by-step Solutions
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Q: Find the value of k for which the function f(x) = kx^2 + 2x + 1 is differentiable at x = 0.
Solution:
f'(x) = 2kx + 2. At x = 0, f'(0) = 2. The function is differentiable for any k, but k = 0 gives a constant function.
Steps: 6
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Step 1: Understand that a function is differentiable at a point if it has a derivative at that point.
Step 2: Write down the function given: f(x) = kx^2 + 2x + 1.
Step 3: Find the derivative of the function, which is f'(x). The derivative of kx^2 is 2kx, and the derivative of 2x is 2. So, f'(x) = 2kx + 2.
Step 4: Evaluate the derivative at x = 0. Substitute 0 into f'(x): f'(0) = 2k(0) + 2 = 2.
Step 5: Since f'(0) = 2, this means the derivative exists and is equal to 2 at x = 0.
Step 6: Conclude that the function is differentiable at x = 0 for any value of k, but if k = 0, the function becomes a constant function.
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