Y=x+√x.x x_{0} = 1

It seems like you are looking at the function \( y = x + \sqrt{x \cdot x} \). Let's break it down and analyze it, especially at the point where \( x_0 = 1 \). Firstly, the expression inside the square root can be simplified since \( \sqrt{x \cdot x} = \sqrt{x^2} = |x| \). For \( x \geq 0 \) (which includes \( x_0 = 1 \)), we have: \[ \sqrt{x^2} = x \] Thus, the function simplifies to: \[ y = x + x = 2x \] Now, substituting \( x_0 = 1 \): \[ y(1) = 2 \cdot 1 = 2 \] So, at \( x_0 = 1 \), the value of \( y \) is 2. If you need more analysis or to find the derivative, limit, or any other specific aspect related to this function at \( x_0 = 1 \), please clarify!