For f(x) = { x^2, x < 1; 2x - 1, x ≥ 1 }, is f differentiable at x = 1?
Practice Questions
1 question
Q1
For f(x) = { x^2, x < 1; 2x - 1, x ≥ 1 }, is f differentiable at x = 1?
Yes
No
Only left
Only right
f'(1) from left = 2, from right = 2; hence f is differentiable at x = 1.
Questions & Step-by-step Solutions
1 item
Q
Q: For f(x) = { x^2, x < 1; 2x - 1, x ≥ 1 }, is f differentiable at x = 1?
Solution: f'(1) from left = 2, from right = 2; hence f is differentiable at x = 1.
Steps: 5
Step 1: Identify the function f(x). It is defined in two parts: f(x) = x^2 when x is less than 1, and f(x) = 2x - 1 when x is greater than or equal to 1.
Step 2: Find the left-hand derivative at x = 1. This means we will use the part of the function where x < 1, which is f(x) = x^2. The derivative of x^2 is 2x. So, we calculate f'(1) from the left: f'(1) = 2(1) = 2.
Step 3: Find the right-hand derivative at x = 1. This means we will use the part of the function where x ≥ 1, which is f(x) = 2x - 1. The derivative of 2x - 1 is 2. So, we calculate f'(1) from the right: f'(1) = 2.
Step 4: Compare the left-hand and right-hand derivatives. Both are equal to 2.
Step 5: Since the left-hand derivative and right-hand derivative at x = 1 are equal, we conclude that f is differentiable at x = 1.
