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 $(0,1)$ rather than $(-\mathrm{\infty },0]\cup [1,\mathrm{\infty })$.

We know from the question that $f(x)=e^x$ when $x\le 0$. Since the exponential function $e^x$ is an increasing function, we have $f(x)\le f(0)=e^0=1$ when $x\le 0$. On the other hand, $e^{-x}$ is decreasing, and hence so is $e^{-(x-1)}=e\cdot e^{-x}$. This follows that $e^{-(x-1)}\ge e^1=e$ when $x\le 0$. This implies that for $x\le 0$, we have

$$f(x)\le 1<e\le e\cdot e^{-x}$$.

(Note that $e\sim 2.7>1$.) In other words, $f(x)\ne e^{-x}$ for any point in $(-\mathrm{\infty },0]$.

In a similar manner, we can show that when $x\ge 1$, we have that $$f(x)=x^2+1\ge 1^2+1=2\text{ but }{e}^{-(x-1)}\le e^{-(1-1)}=1$$ i.e., $f(x)>e^{-(x-1)}$ on $x\in [1,\mathrm{\infty })$

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 $F(x)$ is continuous in the interval $(0,1)$, and

$$F(0)=-k+2-e,F(1)=0+2-1=1.$$

To make $F$ have at least a zero in $(0,1)$, by the theorem, we need to make sure that $F(0)$ and $F(1)$ have opposite signs. i.e., $F(0)\cdot F(1)=(-k+2-e)\cdot 1<0$.

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 $k>2-e.$