The intermediate value theorem states that for a continuous function defined on interval [a,b][a,b] we can let cc be a number with
f(a) < c < f(b) f(a)<c<f(b) and that EE x in [a,b]∃x∈[a,b] such that f(x) = cf(x)=c.
A corollary of this is that if the sign of f(a) !=f(a)≠ sign of f(b)f(b) this means that there must be some x in [a,b]x∈[a,b] such that f(x) = 0f(x)=0 because 00 is obviously between the negatives and positives.
So, let's sub in the endpoints:
f(0) = 0^3 + 0 -1 = -1f(0)=03+0−1=−1
f(1) = 1^3+1 - 1 = 1f(1)=13+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.
