Science:Math Exam Resources/Courses/MATH110/December 2017/Question 09

MATH110 December 2017

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Question 09
Use the Intermediate Value Theorem to show that there is a number ${\textstyle x=c}$ in the interval ${\textstyle (0,2)}$ such that the tangent line to the graph of ${\textstyle f(x)={\frac {1}{4}}x^{4}+x}$ at ${\textstyle x=c}$ is parallel to the tangent line to the graph of ${\textstyle g(x)={\frac {1}{6}}x^{6}}$ at ${\textstyle x=c}$. Is ${\textstyle c}$ closer to 0 or 2?

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
Recall that the Intermediate Value Theorem states that any value, between the functional values on the two ends of an interval, is attained by the function at some point inside the interval.

Hint 2
For two tangent lines to be parallel, the slopes (that is, the derivatives of the functions) must be equal.

Hint 3
We need to find a solution ${\textstyle c}$ to the equation ${\textstyle f'(x)=g'(x)}$ in the interval ${\textstyle (0,2)}$.

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. By the Hints, we need to solve the equation ${\textstyle f'(x)=g'(x)}$ on the interval ${\textstyle (0,2)}$. First we compute $$f'(x)=x^{3}+1,$$ $$g'(x)=x^{5}.$$ So we need to find a root of the equation $$x^{3}+1=x^{5},$$ or $$x^{5}-x^{3}-1=0.$$ Let ${\textstyle F(x)=x^{5}-x^{3}-1}$. In order to apply the Intermediate Value Theorem, we want to check that ${\textstyle 0}$ is between ${\textstyle F(0)}$ and ${\textstyle F(2)}$. Indeed, $$F(0)=0^{5}-0^{3}-1=0-0-1=-1<0,$$ $$F(2)=2^{5}-2^{3}-1=32-8-1=23>0.$$ Therefore ${\textstyle F(0)<0<F(2)}$. Since ${\textstyle F(x)}$ is continuous, by the Intermediate Value Theorem, there exists a ${\textstyle c}$ lying on the interval ${\textstyle (0,2)}$ such that ${\textstyle F(c)=0}$. In other words, there is a point at which ${\textstyle f(x)}$ and ${\textstyle g(x)}$ have parallel tangent lines. In fact, $$F(1)=1^{5}-1^{3}-1=-1<0,$$ there exists a point on $(1,2)$ at which $F$ vanishes. We can consider this point as $c$. Then, $c$ is closer to $2$. To make sure that there's no solution to $F(x)=0$ on $[0,1]$, consider $F'(x)$ on $(0,1)$. Since $F'(x)=4x^{4}-3x^{2}=x^{2}(4x^{2}-3)$, $F$ is decreasing on $(0,{\dfrac {\sqrt {3}}{2}})$ and increasing on $({\dfrac {\sqrt {3}}{2}},1)$. Therefore $F(x)\leq F(0)=-1$ on $(0,{\dfrac {\sqrt {3}}{2}})$ and $F(x)\leq F(1)=-1$ on $({\frac {\sqrt {3}}{2}},1)$. In other words, there's no solution to $F(x)=0$ on $[0,1]$. Answer: $c$ is closer to ${\textstyle {\color {blue}2.}}$