# How do you graph f(x)=-2(3^(x+1))+2 and state the domain and range?

Mar 31, 2018

Domain $\left\{x \in \mathbb{R}\right\}$
Range $\left\{y \in \mathbb{R} | y < 2\right\}$

#### Explanation:

For the domain we are looking for what $x$ cannot be we can do that by breaking down the functions and seeing if any of them yield a result where x is undefined

$u = x + 1$

With this function x is defined for all $\mathbb{R}$ on the number line i.e. all numbers.

$s = {3}^{u}$

With this function u is defined for all $\mathbb{R}$ as u can be negative, positive or 0 without a problem. So through transitivity we know that x is also defined for all $\mathbb{R}$ or defined for all numbers

Lastly

$f \left(s\right) = - 2 \left(s\right) + 2$

With this function s is defined for all $\mathbb{R}$ as u can be negative, positive or 0 without a problem. So through transitivity we know that x is also defined for all $\mathbb{R}$ or defined for all numbers

So we know that x is also defined for all $\mathbb{R}$ or defined for all numbers

$\left\{x \in \mathbb{R}\right\}$

For the range we have to look at what the y values will be for the function

$u = x + 1$

With this function we that there is no value on the number line that won't be u. I.e. u is defined for all $\mathbb{R}$.

$s = {3}^{u}$

With this function we can see that if we place in all the positive numbers $s = {3}^{3} = 27$ we get out another positive number.

Whilst if we place in a negative number $s = {3}^{-} 1 = \frac{1}{3}$ we get a positive number so y can't be negative and will also never be but will approach 0 at $- \infty$

$\left\{s \in \mathbb{R} | s > 0\right\}$

Lastly

$f \left(s\right) = - 2 \left(s\right) + 2$

We see that there is no value $f \left(s\right)$ can equal any value if we disregard what $s$ and $u$ actually state.

But when we look carefully and we consider what $s$ can actually be i.e. only greater than 0. We know that this will effect our final range, as what we see is that every $s$ value is moved up 2 and stretched by -2 when it is placed on the y axis.

So all of the values in s become negative $\left\{f \left(s\right) \in \mathbb{R} | f \left(s\right) < 0\right\}$

Then we know that every value is moved up two

$\left\{f \left(s\right) \in \mathbb{R} | f \left(s\right) < 2\right\}$

so as $f \left(x\right) = f \left(s\right)$ we can say the range is every y value lower than 2
or
$\left\{f \left(x\right) \in \mathbb{R} | f \left(x\right) < 2\right\}$

graph{-2(3^(x+1))+2 [-10, 10, -5, 5]}