Hey guys,

Here's an exercise I was looking at today.

Use the Upper Bound Theorem to find an integral upper bound and the Lower Bound Theorem to find an integral lower bound of the zeros of $\displaystyle f(x)=x^3+3x^2-5x-10$

Here was my explanation:

According to the Upper Bound Theorem, c is an upper bound of the zeros of f(x) if

when you divide f(x) by (x - c), there is no sign change in the quotient or remainder.

So I picked a few possible factors of f(x). First I used (x - 1).

The depressed polynomial produced was $\displaystyle x^2+4x-1-\frac{11}{x-1}$

There is a sign change so 1 is not an upper bound.

Next, I tried (x-2). The depressed polynomial was $\displaystyle x^2+5x+5$

There was no sign change in the quotient or remainder so 2 is an upper bound.

I contend that 2 is the greatest upper bound and there are no zeros of f(x) greater than 2.

Now on to the lower bound.

According to the Lower Bound Theorem, if c is an upper bound of the zeros of f(-x),

then -c is a lower bound of the zeros of f(x).

$\displaystyle f(-x)=-x^3+3x^2+5x-10$

I went through the same process to find the upper bound for f(-x) as I did for f(x).

Turns out that (x - 2) doesn't yield a quotient and remainder with no sign changes.

Neither does (x - 3) or (x - 4). However, (x - 5) does.

This means that all real zeros of f(x) can be found in the interval $\displaystyle -5 \leq x \leq 2$

I agree with this, but is -5 the greatest lower bound?

If you solve the thing algebraically, the roots are $\displaystyle \{\frac{-5 \pm \sqrt{5}}{2}, 2}\}$

This shows the greatest (integral) lower bound is -4.

Shouldn't all the zeros of f(x) be found on this interval: $\displaystyle -4 \leq x \leq 2 \:\:\text{?}$

Since the exercise did not specify the words "greatest" or "least",

maybe they were just looking for any upper and lower bound.

I don't see the significance of that, though.

If that were the case, the interval could be betwen -infinity and +infinity.