Problem 55 - College Algebra and Calculus: An Applied Approach

Question

Describe the \(x\) -values at which the function is differentiable. Explain your reasoning. $$ y=\sqrt{x-1} $$

Step-by-Step Solution

Answer

The function \(y=\sqrt{x-1}\) is differentiable for all \(x\) such that \(x\geq1\).

Step 1: Understand the given function

Given a function \(y=\sqrt{x-1}\) which is a standard square root function. A square root function in general is defined and differentiable for all values where the expression under the square root is non-negative. In this case, the expression under the square root is \(x-1\).

Step 2: Define the domain of the function

The value under the square root should be greater than or equal to 0. That is, \(x-1\geq0\). If we solve for \(x\) we get \(x\geq1\). So the function is defined for all \(x\) that are greater than or equal to 1. Beyond this point the function is continuous and hence differentiable.

Step 3: Conclusion

The function \(y=\sqrt{x-1}\) is differentiable for all \(x\) such that \(x\geq1\).