Hello, geton!
2. a) Sketch the graph of: $\displaystyle y \:=\:e^{\text{-}x} - 1$
On the same axes, sketch the graphs of: .$\displaystyle \begin{array}{cc}y \:= \:\frac{1}{2}(x-1) & \text{for }x \geq 1 \\ y \:= \text{-}\frac{1}{2}(x - 1)&\text{for }x < 1\end{array}$
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 y-axis.
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 x-axis.
The intercepts are: .$\displaystyle \
0,\,0),\;(1,\,0),\;\left(0,\,\frac{1}{2}\right)$
The x-coordinate 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: .$\displaystyle y \:=\:e^{\text{-}x} - 1\text{ and }y \:=\:-\frac{1}{2}(x-1)$
We have: .$\displaystyle e^{\text{-}x} - 1 \:=\:-\frac{1}{2}(x-1)\quad\Rightarrow\quad 2e^{\text{-}x} - 2 \:=\:-x + 1$
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 $\displaystyle x = \text{-}1:\;\begin{array}{ccccc}y & = & e^{\text{-}(\text{-}1)} - 1 & \approx & 1.7 \\ y & = & \text{-}\frac{1}{2}(\text{-}1-1) & = & 1\end{array}$ . . . the exponential is above the line
At $\displaystyle x = 0:\;\begin{array}{ccccc}y & = & e^0-1 &=&0 \\ y&=&\text{-}\frac{1}{2}(0-1) & = & \frac{1}{2}\end{array}$ . . . the exponential is below the line
Therefore, they intersect somewhere on the interval $\displaystyle (\text{-}1,\,0).$