# How do you graph, find the intercepts and state the domain and range of f(x)=2-2^x?

Jun 8, 2018

Refer points below.

#### Explanation:

1. We have to find out the Zeroes of the $f \left(x\right)$ .
$f \left(x\right) = 2 - {2}^{x} = 0$
$\implies$${2}^{x} = 2$
$\implies$$x = 1$
We have one point on the graph is $\left(1 , 0\right)$
2. First Derivative Test:
We have find first derivative of $f \left(x\right)$.
$f \left(x\right) = 2 - {2}^{x}$
$f ' \left(x\right) = - {2}^{x} \ln 2$
Note that $f ' \left(x\right)$ is always negative as${2}^{x} > 0$ and $\ln 2 > 0$.
$f ' \left(x\right) < 0$ for $\forall x \in \mathbb{R}$
$\implies$ $f \left(x\right)$is always deceasing.-------(By First Derivative Test).
3. Second Derivative Test:
We have,
$f \left(x\right) = 2 - {2}^{x}$
$f ' \left(x\right) = - {2}^{x} \ln 2$
$f ' ' \left(x\right) = - {2}^{x} {\left(\ln 2\right)}^{2}$
Note that $f ' ' \left(x\right)$ is always negative.
$f ' ' \left(x\right) < 0$ for $\forall x \in \mathbb{R}$
$\implies$ $f \left(x\right)$is always concave downward.-------(By second Derivative Test).
4. Points on co-ordinate axes:
We have already a point on $x$-axis i.e.$\left(1 , 0\right)$ (as stated in point No. 1).
We have to find out a point on $y$-axis.
So, put $x = 0$ in $f \left(x\right)$
$f \left(x\right) = 2 - {2}^{x}$
$\therefore$$f \left(0\right) = 2 - {2}^{0}$
$\therefore$$f \left(0\right) = 2 - 1$
$\therefore$$f \left(0\right) = 1$
5. Domain:
Clearly there is not any point for which $f \left(x\right)$ is not defined.
Hence, ${D}_{f} = \left(- \infty , \infty\right)$
6. Range:
$f \left(x\right)$ cannot exceed the $y = 2$. Also, $f \left(x\right)$ cannot achieve $y = 2$
Hence, ${R}_{f} = \left(- \infty , 2\right)$
7. Intercepts:
$x$-intercept= 1 unit
$y$-intercept= 1 unit
We have found it by point No.4.
8. Graph:
By above Points we are able to draw the graph.
graph{2-2^x [-10, 10, -5, 5]}