
Patrick B. answered 04/18/20
Math and computer tutor/teacher
f(1) = 1^5 - 3(1) = 1-3 = -2 <0
f(2) = 2^5 - 3(2) = 32-6 = 26 >0
IVT states that there must be a value where f(x) = 0, as this is a continuous polynomial
function on the interval
the approximate solution is by the way 1.316 per the graph