Hi there!
[tex]\large\boxed{\text{No, it does not.}}[/tex]
[tex]y = \frac{x^{2} }{x-1}[/tex]
Looking at the equation, we can see that there is a discontinuity at x = 1 (Vertical Asymptote).
The interval given also includes x = 1 as an enclosed value.
Therefore, the MVT does not apply because the equation is not continuous on the interval [-1, 1].
