graph y = 5^x and y = log(5)^x on a sheet of paper using the same set of axes. Use the graph to describe the domain and range of each function. Then identify the y- intercept of each function and any asymptotes of each function. I already have the graph but idk how to do the rest..
This is the graph
@Owlcoffee Alright, I got it now
Okay, have you tried indentifying the domain of each function? What x-values do each of them exclude?
Like which values the don't land on? @Owlcoffee
I REALLY DONT KNOW LEAVE ME ALONE K? BYE
Sorry? @carlyleukhardt
Yea, look at the curve of e^x. Every single point above the x-axis is a point that corresponds to a x value.
So like -1 and 0 @Owlcoffee
Don't guess, use your knowledge of what "domain of a function" implies and then try to logically draw a conclusion.
I can only think of 1 @Owlcoffee
The Domain is not a specific point, but the restrictions inside a function. For example with the (log (5)x) there are values of x we are not allowed to take.
Then how do I find the domain?
Well, not that is calculable, you have to look at the function and observe which values of x could potentially create an indetermination on the function. For example: \[f(x)=5^x\] Does not have any problems with any value of "x" so we say that the domain are "all the real values".
Ok, and the range wouldn't it be like y>0?
@Owlcoffee
No, you write like this: \[d(f)= \mathbb{R}\] This means that the domain for the function is any real number. And when we speak about domain, we strictly talk about the x-axis.
And I can plug any point on the x axis into that right?
that's correct, that being the case for the first function. What about the other one?
Why can't I do the same for both?
Because they are different function and each has their restrictions. for th case of the first we saw that there is no restrictions, we don't know about the second. Since it's a logarithm you should already know how they go about.
Ok, so the range should be y>0
Yes. But what about the domain?
All real x values right?
No, it's a logarithmic function, it does have restrictions.
oh ok so x>0
Correct. That being the domain for the second function.
Alright, I think I got it. Thank you so much!
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