which x-values at which f is differentiable? f(x)=sqrt (x+1)+1

Answer 1

Hi Michelle,

To see where f is differentiable, we have to take the derivative of the function. Recall that the derivative of √(x) is 1 / (2√(x)), and that the derivative of a constant is 0. So, the derivative of f(x) is

f'(x) = 1 / (2√(x + 1)).

The domain of f'(x) is the same as where f(x) is differentiable. To find the domain, we have to look at the limitations of where we can substitute x. For f'(x), this is the square root and the fraction. The part under the square root (x + 1) has to be greater than or equal to 0. As an inequality, this is x + 1 ≥ 0, or x ≥ -1. Also, the denominator of f'(x) cannot be 0, or 2√(x + 1) ≠ 0. Rearranging this inequality for x, we get x ≠ -1. Combining the inequalities from the square root (x ≥ -1) and the denominator (x ≠ -1), we get the domain of f'(x) to be x > -1 (changed the "greater than or equal to" to just "greater than"), and this is where f(x) is differentiable.