If , then
- A.is continuous but not differentiable at
- B.is differentiable at
- C.is not differentiable at
- D.None of the above
Show answer and solution
Correct option: C
Solution
Rationalize: f(x) = x(√x + √(x+1)) = x·[(x+1) - x]/(√(x+1) - √x) ... actually let's compute f'(0) directly.
f(0) = 0. For x > 0, f'(0) = lim(x→0⁺) f(x)/x = lim(x→0⁺) (√x + √(x+1)) = 0 + 1 = 1.
For x < 0 (approaching from the left), f(x) = x(√x + √(x+1)) is only defined when both √x and √(x+1) are real, requiring x ≥ 0 and x ≥ -1. So the domain is x ≥ 0, and x = 0 is a boundary point. The left-hand derivative doesn't exist since f is not defined for x < 0. Therefore f(x) is not differentiable at x = 0.
f(0) = 0. For x > 0, f'(0) = lim(x→0⁺) f(x)/x = lim(x→0⁺) (√x + √(x+1)) = 0 + 1 = 1.
For x < 0 (approaching from the left), f(x) = x(√x + √(x+1)) is only defined when both √x and √(x+1) are real, requiring x ≥ 0 and x ≥ -1. So the domain is x ≥ 0, and x = 0 is a boundary point. The left-hand derivative doesn't exist since f is not defined for x < 0. Therefore f(x) is not differentiable at x = 0.
[1985]