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

MATH110 December 2011

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Question 03
Prove that the function $\displaystyle f(x)=x^{3}-15x+1$ has three roots in the interval [-4, 4]. Make sure to state any assumptions you are making, or theorems you are using.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Can you see the shape of a cubic polynomial having three roots? What can you use to prove that the given polynomial has that shape without having to find out what the roots are?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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. The three intervals are contained in [-4,4]. The three intervals don't overlap. 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.

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