Science:Math Exam Resources/Courses/MATH110/December 2014/Question 11

MATH110 December 2014

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Question 11
Show that there is at least one point on the curve ${\textstyle y=x^{4}+2x^{2}-4x-1}$ where the tangent line to the curve is horizontal.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The slope of the tangent line of the given curve at a point $x=a$ is ${\frac {dy}{dx}}{\bigg |}_{x=a}$.

Hint 2
Use the intermediate value theorem.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. Since the slope of the tangent line at a point $x=a$ is $y'|_{x=a}$, to find the horizontal tangent line, it is enough to solve $y'=0$. Indeed, if we find a point $a$ satisfying $y'(a)=0$, the tangent line of the given curve at $x=a$ is horizontal. First, we find the derivative using the power rule; ${\frac {dy}{dx}}=y'=(x^{4}+2x^{2}-4x-1)'=4x^{3}+4x-4=4(x^{3}+x-1).$ Now, the question is reduced to find a solution to $g(x)=x^{3}+x-1=0$. For this purpose, we use the intermediate value theorem; Observe that $g(x)=x^{3}+x-1$ is a polynomial, and hence it is continuous for any point in the real line. On the other hand, by plugging some numbers, we get $g(0)=-1<0$, $g(1)=1+1-1=1>0$. Then, by the intermediate value theorem, we can find a point $a\in (0,1)$ such that $g(a)=0$. In other words, there exists a point $a$ such that $y'(a)=0$, so that at that point, the corresponding tangent line is horizontal.