+0  
 
0
1029
4
avatar+253 

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

 Jun 27, 2014

Best Answer 

 #4
avatar+109783 
+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 50=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

 Jun 28, 2014
 #1
avatar+111396 
+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??

 

 Jun 27, 2014
 #2
avatar+253 
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.

 Jun 27, 2014
 #3
avatar+111396 
+5

OK....

f(x) = 5-x   can be wrtten as  1/5x

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 !!!

 

 Jun 27, 2014
 #4
avatar+109783 
+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 50=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

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