Science:Math Exam Resources/Courses/MATH110/December 2016/Question 10

MATH110 December 2016

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Question 10
Show that the tangent line to the curve $y=x^{3}-x+3$ at the point $(1,y(1))$ intersects the curve $y=e^{x}$ twice.

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 tangent line of a function ${\textstyle y=f(x)}$ at the point ${\textstyle x=a}$ is given by ${\textstyle y=f(a)+f'(a)(x-a)}$.

Hint 2
Two curves (including lines), say intersect if there are ${\textstyle (x,y)}$ satisfying both equations.

Hint 3
In order to show that some equation ${\textstyle F(x)=0}$ have a solution, one can use the intermediate value theorem and check that such function ${\textstyle F(x)}$ is positive here and negative there.

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 Hint 1, the tangent line to the curve ${\textstyle y=x^{3}-x+3}$ at the point ${\textstyle (1,y(1))}$ is $$y=y(1)+y'(1)(x-1).$$ Clearly, $$y(1)=1^{3}-1+3=3.$$ Also, since $$y'(x)=3x^{2}-1,$$ at ${\textstyle x=1}$ we have $$y'(1)=3\cdot 1^{2}-1=3-1=2.$$ This gives the equation of the tangent line, $$y=3+2(x-1),$$ that is, $$y=2x+1.$$ We need to check that the tangent line ${\textstyle y=2x+1}$ intersects the curve ${\textstyle y=e^{x}}$ twice. In other words, according to Hint 2, we need to find an ${\textstyle x}$ such that $$2x+1=e^{x}.$$ (In such case, ${\textstyle y}$ is automatically determined by either ${\textstyle y=2x+1}$ or ${\textstyle y=e^{x}}$.) Therefore, if we write $$F(x)=e^{x}-2x-1,$$ then it reamins to solve the equation $$F(x)=0.$$ We observe that $$F(0)=e^{0}-2\cdot 0-1=1-0-1=0.$$ This means that the tangent line ${\textstyle y=2x+1}$ intersects the curve ${\textstyle y=e^{x}}$ at ${\textstyle x=0}$ (and ${\textstyle y=1}$). By Hint 3, let us use the intermediate value theorem to find another intersection point. We need to find points where ${\textstyle F(x)}$ is positive and points where it is negative. Observe that $${\begin{aligned}F(1)&=e^{1}-2\cdot 1-1=e-2-1=e-3<0,\\F(2)&=e^{2}-2\cdot 2-1=e^{2}-4-1=e^{2}-5>0.\end{aligned}}$$ Here, we have used the fact that ${\textstyle e\approx 2.7}$, so that ${\textstyle {\sqrt {5}}<e<3}$, that is, ${\textstyle e<3}$ and ${\textstyle e^{2}>5}$. Since ${\textstyle F(x)}$ is continuous everywhere, by intermediate value theorem, we conclude that ${\textstyle F(c)=0}$ for some ${\textstyle 1<c<2}$. In summary, the tangent line ${\textstyle y=2x+1}$ intersect the curve ${\textstyle y=e^{x}}$ at ${\textstyle x=0}$ and at a point ${\textstyle c}$ between ${\textstyle 1}$ and ${\textstyle 2}$. Answer: ${\textstyle {\color {blue}{\mbox{The tangent line is found to be }}y=2x+1.}}$ ${\textstyle {\color {blue}{\mbox{ By inspection, it intersects the curve }}y=e^{x}{\mbox{ at the point where }}x=0.}}$ ${\textstyle {\color {blue}{\mbox{By the intermediate value theorem, they intersect at a point in the interval }}(1,2).}}$