Is ##f(x)=2^{x}-1## considered an exponential function?

[Callmelucky]

Homework Statement

I wonder if it's $f(x)=2^{x}-1$ considered an exponential function?

Relevant Equations

$f(x)=a^{x}$

I wonder if it's $f(x)=2^{x}-1$ considered an exponential function because in my textbook it's stated that the set of values of an exponential function is a set of positive real numbers, while when graphing this function I get values(y line) that are not positive(graph in attachments), so I am not sure if the statement "the set of values of an exponential function is a set of positive real numbers" applies only to this type of function $f(x)=a^{x}$ or does it include all types of functions, like this one $f(x)=2^{x}-1$?

Thank you

 

[Mark44]

Homework Statement:: I wonder if it's $f(x)=2^{x}-1$ considered an exponential function?
Relevant Equations:: $f(x)=a^{x}$

I wonder if it's $f(x)=2^{x}-1$ considered an exponential function because in my textbook it's stated that the set of values of an exponential function is a set of positive real numbers, while when graphing this function I get values(y line) that are not positive(graph in attachments), so I am not sure if the statement "the set of values of an exponential function is a set of positive real numbers" applies only to this type of function $f(x)=a^{x}$ or does it include all types of functions, like this one $f(x)=2^{x}-1$?

Thank you

Yes, this is an exponential function. Your textbook is considering only functions of the form $f(x) = a^x$, which would have only positive values. The one you asked about is the translation down by 1 unit of $y = 2^x$, so the translated version will have negative values when x < 0.

 

[fresh_42]

Homework Statement:: I wonder if it's $f(x)=2^{x}-1$ considered an exponential function?
Relevant Equations:: $f(x)=a^{x}$

I wonder if it's $f(x)=2^{x}-1$ considered an exponential function because in my textbook it's stated that the set of values of an exponential function is a set of positive real numbers, while when graphing this function I get values(y line) that are not positive(graph in attachments), so I am not sure if the statement "the set of values of an exponential function is a set of positive real numbers" applies only to this type of function $f(x)=a^{x}$ or does it include all types of functions, like this one $f(x)=2^{x}-1$?

Thank you

This depends more on the context than on a precise definition.

as an algebraic object:
a linear combination of an exponential ($x\mapsto 2^x$) and a constant ($x\mapsto 1$) function

as an algorithmic runtime:
an exponential function, the shift by $-1$ is irrelevant

as an analytical function:
a shifted (by $c$) exponential function ($x\mapsto a^x+c$)

It is not purely an exponential function, but the effect of minus one is in almost all cases negligible so people might call it exponential despite of it.

 

[Callmelucky]

Yes, this is an exponential function. Your textbook is considering only functions of the form $f(x) = a^x$, which would have only positive values. The one you asked about is the translation down by 1 unit of $y = 2^x$, so the translated version will have negative values when x < 0.

thank you

 

[Callmelucky]

This depends more on the context than on a precise definition.

as an algebraic object:
a linear combination of an exponential ($x\mapsto 2^x$) and a constant ($x\mapsto 1$) function

as an algorithmic runtime:
an exponential function, the shift by $-1$ is irrelevant

as an analytical function:
a shifted (by $c$) exponential function ($x\mapsto a^x+c$)

It is not purely an exponential function, but the effect of minus one is in almost all cases negligible so people might call it exponential despite of it.

thank you

 

[WWGD]

Ultimately, for $a>0$, you may write : $a^x =e^{x ln(a)},$ so I'd say it qualifies.

 

[FactChecker]

When considering the growth of the function for large x, I would call it exponential growth. But IMO, to call it an exponential function is a mistake. There are too many situations where you would have to continually mention the added constant.

 

[WWGD]

When considering the growth of the function for large x, I would call it exponential growth. But IMO, to call it an exponential function is a mistake. There are too many situations where you would have to continually mention the added constant.

Still, as $a$ grows, the value of the function and it's translate will become very close, even if the ln slows the growth of the $a$

 

[malawi_glenn]

Similar question: is $f(x) = kx + m$, $(m \neq 0 )$ considered to be a linear function?

It depends on your definition.
Some say "yes" (usually in calculus) because the graph is a straight line.
Some say "no" (usually in linear algebra) because it does not fulfill $f(x_1 + x_2) = f(x_1) + f(x_2)$ and $f(ax) = af(x)$.

 

[Mark44]

Yes, this is an exponential function.

Maybe not an exponential function per se, but definitely a simple transformation of one.

 

[FactChecker]

Maybe not an exponential function per se, but definitely a simple transformation of one.

Agreed. The only reason I can think of to not call it officially an exponential function is this. If there are theorems about exponential functions, they might not apply to this function.

 

[fresh_42]

Maybe not an exponential function per se, but definitely a simple transformation of one.

It depends on what we consider the essential information and it therewith depends on context. I am used to complexity considerations so $f(x)=O(2^x).$ Others may consider them as linear independent functions in some algebra, $2^x$ and $-1.$ Again others may see its asymptotic behavior, i.e. the exponential part.

The question becomes more interesting if we consider examples like $f(x)=2^x+x^2+x \log x +c.$ Would we still call it exponential? Probably not, although it is still $f(x)=O(2^x).$ So that would be a non-exponential function with an exponential behavior.

 

[WWGD]

Similar question: is $f(x) = kx + m$, $(m \neq 0 )$ considered to be a linear function?

It depends on your definition.
Some say "yes" (usually in calculus) because the graph is a straight line.
Some say "no" (usually in linear algebra) because it does not fulfill $f(x_1 + x_2) = f(x_1) + f(x_2)$ and $f(ax) = af(x)$.

That's a(ffine) example you used.