Observe that
"f(-3)=(-3)^3-3(-3)+5=-13<0""f(-1)=(-1)^3-3(-1)+5=7>0"
Since "f" is continuous, we may conclude by the IVT that "f" has a root in "[-3, -1]."
Now
So "f(-3)" and "f(-2)" are of opposite sign. Therefore, the IVT guarantees that "f" has a root on "[-3, -2]."
We find the interval of length 1 containing a root or solution of "f(x)=x^3 - 3x + 5."
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