The function f(x) = sqrt(x) is differentiable at x = 0?
Practice Questions
1 question
Q1
The function f(x) = sqrt(x) is differentiable at x = 0?
Yes
No
Only from the right
Only from the left
f(x) = sqrt(x) is not differentiable at x = 0 because the left-hand derivative does not exist.
Questions & Step-by-step Solutions
1 item
Q
Q: The function f(x) = sqrt(x) is differentiable at x = 0?
Solution: f(x) = sqrt(x) is not differentiable at x = 0 because the left-hand derivative does not exist.
Steps: 7
Step 1: Understand what it means for a function to be differentiable at a point. A function is differentiable at a point if it has a defined derivative at that point.
Step 2: Recall the function given: f(x) = sqrt(x). This function is defined for x >= 0, meaning we can only consider values of x that are 0 or positive.
Step 3: To check if f(x) is differentiable at x = 0, we need to find the derivative at that point. The derivative can be found using the limit definition of the derivative.
Step 4: The derivative of f(x) at x = 0 is given by the limit: f'(0) = lim (h -> 0) [(f(0 + h) - f(0)) / h]. Since f(0) = sqrt(0) = 0, this simplifies to: f'(0) = lim (h -> 0) [sqrt(h) / h].
Step 5: Simplifying sqrt(h) / h gives us 1/sqrt(h). As h approaches 0 from the right (positive side), 1/sqrt(h) approaches infinity, which means the limit does not exist.
Step 6: Since the limit does not exist, the right-hand derivative at x = 0 is infinite, and we cannot find a left-hand derivative because the function is not defined for negative values of x.
Step 7: Therefore, since the left-hand derivative does not exist and the right-hand derivative is infinite, we conclude that f(x) = sqrt(x) is not differentiable at x = 0.
