Thread: What is the domain of this function? (I am so confused)

Hello.

I would like to ask what is the domain of the following fucnction.

y=(5x^x)^(x+1)

Our teachers tell us that the domain of a power function (A^X where A is constant and X is a variable) is that the base must be positive and must differ from 1,so in the above example, it is logically that the solution of the following system of inequalities defines the domain:

5x>0 and
5x<>1 and
5x^x>0 and
5x^x<>1


Am I right? And I must note that in our textbook the power function's domain(sorry if I misuse mathematical argo) A^X is that the base A must be positive but it does not say that it must differ from 1. Ohhhh! Please clarify this very obscure situation!

Originally Posted by Val21

Hello.

I would like to ask what is the domain of the following fucnction.

y=(5x^x)^(x+1)

Our teachers tell us that the domain of a power function (A^X where A is constant and X is a variable) is that the base must be positive and must differ from 1,so in the above example, it is logically that the solution of the following system of inequalities defines the domain:

5x>0 and
5x<>1 and
5x^x>0 and
5x^x<>1


Am I right? And I must note that in our textbook the power function's domain(sorry if I misuse mathematical argo) A^X is that the base A must be positive but it does not say that it must differ from 1. Ohhhh! Please clarify this very obscure situation!

$\displaystyle y=\left ( 5x^x \right )^{x+1} = 5^{x+1} \cdot x^{x(x+1)}$

Looking at these terms individually, we see that we have no restrictions on the function due to the first factor. The second factor gives us that $\displaystyle x \geq 0$, since the base must be non-negative, but note that if x = 0 the second factor is $\displaystyle 0^0$, which is undefined. So the domain is x > 0.

-Dan

And what if we take X for 1 so we would get number 1 in the base ,which is prohibited. So probably the domain must be x>0 and x<>1 ?

Thanks for replying

Originally Posted by Val21

And what if we take X for 1 so we would get number 1 in the base ,which is prohibited. So probably the domain must be x>0 and x<>1 ?

Thanks for replying

$\displaystyle x$ can also be any integer, so it looks to me as though the domain is:

$\displaystyle
\left[ \mathbb{R}_+ \cup \mathbb{Z} \right] \backslash \{0\}
$

The union of all positive reals, and non-zero integers.

On second thoughts it looks as though this is also defined when $\displaystyle x$ is a rational of
the form $\displaystyle -\frac{2n}{2m-1},\ n,\ m \in \mathbb{Z}_+$ (negative rationals which in lowest form have an even numerator)

So my latest bet is:

$\displaystyle
\left[ \mathbb{R}_+ \cup \left( \mathbb{Z}\backslash \{0\} \right) \cup A \right]
$

where $\displaystyle A=\{a: a=-2n/(2m-1),\ n,\ m \in \mathbb{Z}_+ \}$.

But somehow I don't think your teacher is looking for an answer of this nature.

RonL

Originally Posted by Val21

And what if we take X for 1 so we would get number 1 in the base ,which is prohibited. So probably the domain must be x>0 and x<>1 ?

Thanks for replying

I don't see why $\displaystyle 1$ should be prohibited in the base, $\displaystyle 0$ yes as $\displaystyle 0^0$ is undefined, but not $\displaystyle 1$.

When $\displaystyle x=1$, we have:

$\displaystyle y=(5x^x)^{x+1}=(5 \times 1^1)^2=5^2=25 $

RonL