Science:Math Exam Resources/Courses/MATH110/December 2011/Question 03/Solution 1

First, we are of course interested in the values of the function at x = -4 and x = 4. We have:

${\begin{aligned}f(-4)&=(-4)^{3}-15(-4)+1=-64+60+1=-3\\f(4)&=4^{3}-15\cdot 4+1=64-60+1=5\\\end{aligned}}$

So, using the Intermediate Value Theorem (since this polynomial is clearly continuous everywhere) we can guarantee the existence of one root. But not three.

To show there are three roots, we need to show that this cubic polynomial crosses the x-axis again. If we just find the local extrema (if there are three roots, there should be one local maximum followed by one local minimum) and show that they are in our interval, we could refine our use of the Intermdeiate Value Theorem.

Let's find out where these extrema are. First, we compute the derivative of the function ƒ:

$\displaystyle f'(x)=3x^{2}-15$

And then find out the critical points:

$\displaystyle f'(x)=0\iff 3x^{2}-15=0\iff x^{2}=5\iff x=\pm {\sqrt {5}}$

We actually only need to know what is the value of the function ƒ at these two points:

${\begin{aligned}f(-{\sqrt {5}})=-5{\sqrt {5}}+15{\sqrt {5}}+1=5{\sqrt {5}}+1>0\\f({\sqrt {5}})=5{\sqrt {5}}-15{\sqrt {5}}+1=-10{\sqrt {5}}+1<0\end{aligned}}$

(If you are not sure how we know that the second value is negative, consider that √5 > 2 so -10√5 < -20).

So we can now apply the Intermediate Value Theorem three times, on the intervals [-4, -√5], [-√5, √5] and [√5, 4]] to show that there must be a root in each of these intervals.

Note that we need not use these intervals precisely. As long as you can identify three intervals similar to the ones above, i.e.

1. The three intervals are contained in [-4,4].
2. The three intervals don't overlap.
3. If for one endpoint of an interval, $f>0$ then for the other endpoint, $f<0$.

Then you can use the intermediate value theorem in the same way as described above to show there is a root in each interval.