The function f(x) = { x^2, x < 1; 2x - 1, x ≥ 1 } is differentiable at x = 1 if which condition holds?
Practice Questions
1 question
Q1
The function f(x) = { x^2, x < 1; 2x - 1, x ≥ 1 } is differentiable at x = 1 if which condition holds?
f(1) = 1
f'(1) = 1
f'(1) = 2
f(1) = 2
For differentiability, the left and right derivatives must equal at x = 1, hence f'(1) = 1.
Questions & Step-by-step Solutions
1 item
Q
Q: The function f(x) = { x^2, x < 1; 2x - 1, x ≥ 1 } is differentiable at x = 1 if which condition holds?
Solution: For differentiability, the left and right derivatives must equal at x = 1, hence f'(1) = 1.
Steps: 5
Step 1: Understand that the function f(x) has two parts: one for x < 1 (which is x^2) and one for x ≥ 1 (which is 2x - 1).
Step 2: To check if f(x) is differentiable at x = 1, we need to find the derivative from the left side (as x approaches 1 from the left) and the right side (as x approaches 1 from the right).
Step 3: Calculate the left derivative at x = 1 using the part of the function for x < 1: f'(x) = 2x. So, f'(1) from the left is 2(1) = 2.
Step 4: Calculate the right derivative at x = 1 using the part of the function for x ≥ 1: f'(x) = 2. So, f'(1) from the right is 2.
Step 5: For f(x) to be differentiable at x = 1, the left derivative (2) must equal the right derivative (2). Since they are equal, f(x) is differentiable at x = 1.
