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 = (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].