Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (c)

MATH110 April 2016

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q5 (e)  •  Q5 (f)  •  Q5 (g)  •  Q5 (h)  •  Q5 (i)  •  Q6 (a)  •  Q6 (b)  •  Q7  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  Q8 (d)  •  Q9 (a)  •  Q9 (b)  •  Q10  •  

   Other MATH110 Exams

  •  December 2011  •  December 2010  •  April 2012  •  April 2011  •  December 2012  •  April 2013  •  December 2013  •  April 2014  •  December 2014  •  December 2015  •  April 2017  •  April 2016  •  December 2016  •  December 2017  •  April 2018  •  

[hide] Question 03 (c)

Consider the following function, where $k$ is a constant. ${\begin{aligned}f(x)={\begin{cases}e^{x}&{\text{if }}x\leq 0\\k(x-1)+2&{\text{if }}0<x<1\\x^{2}+1&{\text{if }}x\geq 1.\end{cases}}\end{aligned}}$ For what value(s) of $k$ does the equation $f(x)-e^{-(x-1)}=0$ have at least one solution? Explain why. Also, make sure you state what theorem(s) you may be using to justify your claim and why they apply.

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.

[show] Hint 1
To show that this equation has zero is equivalent to show that the function ${\textstyle f(x)}$ has intersection with ${\textstyle e^{-(x-1)}}$. Observe that the intersection can only happen in the interval ${\textstyle (0,1)}$.

[show] Hint 2
Apply intermediate value theorem on the interval ${\textstyle (0,1)}$.

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.

[show] Solution

Before we apply intermediate value theorem, let’s discuss why the intersection can only happen in ${\textstyle (0,1)}$ rather than ${\textstyle (-\infty ,0]\cup [1,\infty )}$. We know from the question that ${\textstyle f(x)=e^{x}}$ when ${\textstyle x\leq 0}$. Since the exponential function $e^{x}$ is an increasing function, we have $f(x)\leq f(0)=e^{0}=1$ when ${\textstyle x\leq 0}$. On the other hand, $e^{-x}$ is decreasing, and hence so is $e^{-(x-1)}=e\cdot e^{-x}$. This follows that $e^{-(x-1)}\geq e^{1}=e$ when ${\textstyle x\leq 0}$. This implies that for $x\leq 0$, we have $$f(x)\leq 1<e\leq e\cdot e^{-x}$$ . (Note that $e\sim 2.7>1$.) In other words, $f(x)\neq e^{-x}$ for any point in ${\textstyle (-\infty ,0]}$. In a similar manner, we can show that when ${\textstyle x\geq 1}$, we have that $$f(x)=x^{2}+1\geq 1^{2}+1=2{\text{ but }}\ e^{-(x-1)}\leq e^{-(1-1)}=1$$ i.e., ${\textstyle f(x)>e^{-(x-1)}}$ on ${\textstyle x\in [1,\infty )}$ The above analysis would be more obvious if you draw the graphs of $f$ and $g$. Now, we find the value $k$ which makes $f(x)-e^{-(x-1)}=0$ have at least one solution on the interval $(0,1)$, by using the intermediate value theorem. On the interval $(0,1)$, let $$F(x)=f(x)-e^{-(x-1)}=k(x-1)+2-e^{-(x-1)}.$$ (Here, we use $f(x)=k(x-1)+2$ on $(0,1)$.) It is time to apply intermediate value theorem, note that ${\textstyle F(x)}$ is continuous in the interval ${\textstyle (0,1)}$, and $$F(0)=-k+2-e,\ F(1)=0+2-1=1.$$ To make $F$ have at least a zero in ${\textstyle (0,1)}$, by the theorem, we need to make sure that $F(0)$ and $F(1)$ have opposite signs. i.e., ${\textstyle F(0)\cdot F(1)=(-k+2-e)\cdot 1<0}$. Solving this inequality gives us that $$k>2-e.$$ Thus, we have at least one solution to $F(x)=0$ (i.e., $f(x)=g(x)=e^{-(x-1)}$) if ${\textstyle \color {blue}k>2-e.}$