Hello, geton!
2. a) Sketch the graph of: $\displaystyle y \:=\:e^{\text{}x}  1$
On the same axes, sketch the graphs of: .$\displaystyle \begin{array}{cc}y \:= \:\frac{1}{2}(x1) & \text{for }x \geq 1 \\ y \:= \text{}\frac{1}{2}(x  1)&\text{for }x < 1\end{array}$
Show the coordinates of the points where the graph meets the axes. Code:

* 
* 
* *½ *
*  * *
*  * 1 *
**
0 *
 *
 *
   1+          

We're expected to be familiar with the graph of $\displaystyle (a)\;y \:=\:e^x$
The graph of $\displaystyle (b)\;y \:=\:e^{\text{}x}$ is a reflection of (a) over the yaxis.
The graph of $\displaystyle y\:=\:e^{\text{}x}  1$ is the graph of (b) lowered one unit.
The two lines intersect at (1,0) and exist above the xaxis.
The intercepts are: .$\displaystyle \0,\,0),\;(1,\,0),\;\left(0,\,\frac{1}{2}\right)$
The xcoordinate of the point of intersection of the graphs is $\displaystyle \alpha.$
b) Show that $\displaystyle x = \alpha$ is a root of the equation $\displaystyle x + 2e^{\text{}x}  3 \:= \:0$
The only intersection is that of: .$\displaystyle y \:=\:e^{\text{}x}  1\text{ and }y \:=\:\frac{1}{2}(x1)$
We have: .$\displaystyle e^{\text{}x}  1 \:=\:\frac{1}{2}(x1)\quad\Rightarrow\quad 2e^{\text{}x}  2 \:=\:x + 1$
Therefore, $\displaystyle \alpha$ is a root of: .$\displaystyle x + 2e^{\text{}x}  3 \:=\:0$
c) Show that: .$\displaystyle 1 \,<\,\alpha \,<\,0$
Since both the exponential function and the line are continuous,
. . the most elementary method is to test the endpoints.
At $\displaystyle x = \text{}1:\;\begin{array}{ccccc}y & = & e^{\text{}(\text{}1)}  1 & \approx & 1.7 \\ y & = & \text{}\frac{1}{2}(\text{}11) & = & 1\end{array}$ . . . the exponential is above the line
At $\displaystyle x = 0:\;\begin{array}{ccccc}y & = & e^01 &=&0 \\ y&=&\text{}\frac{1}{2}(01) & = & \frac{1}{2}\end{array}$ . . . the exponential is below the line
Therefore, they intersect somewhere on the interval $\displaystyle (\text{}1,\,0).$