Question

Does the MVT apply? Explain why or why not. y=-x^2/x-1 ; [-1,1]

Answer 1

Hi there!

$\large\boxed{\text{No, it does not.}}$

$y = \frac{x^{2} }{x-1}$

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].