The function f(x) = x^2 for x < 1 and f(x) = 2x - 1 for x ≥ 1 is differentiable at x = 1?

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Q: The function f(x) = x^2 for x < 1 and f(x) = 2x - 1 for x ≥ 1 is differentiable at x = 1?

Solution: f'(1) from left = 2 and from right = 2; hence, f is continuous but not differentiable at x = 1.

Steps: 10

Step 1: Identify the function f(x). It has two parts: f(x) = x^2 for x < 1 and f(x) = 2x - 1 for x ≥ 1.
Step 2: Check if the function is continuous at x = 1. This means we need to find f(1) from both sides.
Step 3: Calculate f(1) using the second part of the function (since x = 1 falls in x ≥ 1). So, f(1) = 2(1) - 1 = 1.
Step 4: Calculate the limit of f(x) as x approaches 1 from the left (x < 1). This is f(1) = 1^2 = 1.
Step 5: Calculate the limit of f(x) as x approaches 1 from the right (x ≥ 1). This is f(1) = 2(1) - 1 = 1.
Step 6: Since both limits equal f(1), the function is continuous at x = 1.
Step 7: Now, find the derivative from the left side (x < 1). The derivative of f(x) = x^2 is f'(x) = 2x, so f'(1) from the left = 2(1) = 2.
Step 8: Find the derivative from the right side (x ≥ 1). The derivative of f(x) = 2x - 1 is f'(x) = 2, so f'(1) from the right = 2.
Step 9: Compare the left and right derivatives. Both are equal to 2.
Step 10: Since the left and right derivatives are equal, the function is differentiable at x = 1.