Would i be able to get the domain and range from a function or does it have to be from a graph?

sally1
Jun 27, 2014

#4**+5 **

CPhill has covered it well but I will stress something here.

I specifically want to look at $$f(x)=5^{-x}$$

If x is negative this becomes

$$5^{--number}=5^{+number}=positive\:\:number$$

If x=0 then 5^{0}=1 = positive number

If x is postitive then

$$5^{-number} = \dfrac{1}{5^{+\:number}}=\dfrac{1}{+\:\: number}=positive\:\: number$$

When the + number on the bottom gets very big this will approach 0 but it won't ever actually get there.

So f(x)=0 is an ASYMTOTE f(x)>0

so the range is $$(0,\infty)\quad \mbox{ or put differently }\quad 0

I have written some posts on negative indices it may be useful for you to revise.

http://web2.0calc.com/questions/indices-especially-negative-indices

Melody
Jun 28, 2014

#1**+5 **

Graphing is sometimes easier, but we can often analyze the function, too.......

Sometimes....I use both....just to verify my answer

Do you have a particular function in mind??

CPhill
Jun 27, 2014

#2**0 **

yes. f(x) = 5^{−x }

I'm thinking that it is -infinity, positive infity for the domain but i have no idea if this is right. and i do not know how to find the range.

sally1
Jun 27, 2014

#3**+5 **

OK....

f(x) = 5^{-x} can be wrtten as 1/5^{x}

Note that we can put any real number in for "x" because an exponential function never = 0 , so the denominator never = 0 So "domain" relates to x....and the domain is just (-∞ , ∞).....just as you suspected........!!!

The range - which relates to "y" - is trickier....as x is more and more negative, the function grows larger and as x gets more positive, the function gets very close to 0 (but not actually 0). So the range is (0 , ∞). Here's where a graph may help!!

Note how the graph gets "close" to 0 on the right, but is unbounded on the left!!

Hope this helps !!!

CPhill
Jun 27, 2014

#4**+5 **

Best Answer

CPhill has covered it well but I will stress something here.

I specifically want to look at $$f(x)=5^{-x}$$

If x is negative this becomes

$$5^{--number}=5^{+number}=positive\:\:number$$

If x=0 then 5^{0}=1 = positive number

If x is postitive then

$$5^{-number} = \dfrac{1}{5^{+\:number}}=\dfrac{1}{+\:\: number}=positive\:\: number$$

When the + number on the bottom gets very big this will approach 0 but it won't ever actually get there.

So f(x)=0 is an ASYMTOTE f(x)>0

so the range is $$(0,\infty)\quad \mbox{ or put differently }\quad 0

I have written some posts on negative indices it may be useful for you to revise.

http://web2.0calc.com/questions/indices-especially-negative-indices

Melody
Jun 28, 2014