The Intermediate Value Theorem states that, if a continuous function over an interval [a,b] is such that f(a)=x and f(b)=y, then f assumes all possible values between x and y in the interval [a,b]. This function is mostly used to find zeroes of a function, by finding a point in which the function is negative, and one in which it's positive: passing from negative to positive (i.e., from below to above the x-axis), the function must cross it, and that's the root.
So, we must verify that f(x) is continuous in [−3,−1], and that f(−3) and f(−1) have opposite signs.
Coefficients apart, the first two terms are power of x, and so they are continuous everywhere. The third term, 1x, has its only discontinuity in x=0, which is outside of our domain [−3,−1], so f is indeed continuous in [−3,−1].
Now we only need to compute:
f(−3)=(−3)32−4⋅(−3)+1−3=−116<0
f(−1)=−12−4(−1)−1=4−1−12=52>0.
