Science:Math Exam Resources/Courses/MATH110/April 2016/Question 03 (c)/Solution 1

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