Mean Value Theorem Practice Test

Yes; because $t(-4)>t(2)$, there must be $c$ with $t'(c)=-1$ even without continuity.

No; MVT requires $t$ to be differentiable at $x=-4$ and $x=2$.

Yes; continuity on $(-4,2)$ alone guarantees a point where $t'(c)=-1$.

Yes; since $t$ is continuous on $[-4,2]$ and differentiable on $(-4,2)$, some $c$ has $t'(c)=\dfrac{4-10}{2-(-4)}=-1$.

No; MVT applies only when the function is linear.