Feel free to do that. Go ahead. State it.

Then notice that:

f(0)= (0^5-0+8)sin(0(pi)) = 8*0 = 0

f(1)= (1^5-1+8)sin(1(pi)) = 8*0 = 0

That's unusual. Is f(x) a constant function? Maybe we should appeal tothe criteria for the Mean Value Theorem.ALL

Note: Please stop posting problems with no work shown. Personally, I weary quickly of such laziness.