The intermediate value theorem states that for a continuous function defined on interval #[a,b]# we can let #c# be a number with
#f(a) < c < f(b) # and that #EE x in [a,b]# such that #f(x) = c#.
A corollary of this is that if the sign of #f(a) !=# sign of #f(b)# this means that there must be some #x in [a,b]# such that #f(x) = 0# because #0# is obviously between the negatives and positives.
So, let's sub in the endpoints:
#f(0) = 0^3 + 0 -1 = -1#
#f(1) = 1^3+1 - 1 = 1#
#therefore# there is at least one zero in this interval. To check if there's only one root we look at the derivative which gives the slope.
#f'(x) = 3x^2 + 1#
We can see that #AA x in [a,b], f'(x) > 0# so the function is always increasing in this interval - this means there is only one root in this interval.
