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

MATH110 April 2016

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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.

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 ${\textstyle (0,1)}$.

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

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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. 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 ${\textstyle \color {blue}k>2-e.}$