Question

HELP PLEASE! ALGEBRA!!!
Regarding y= x^x, write a brief statement on why all positive x-values yield real solutions and why only some negative x-values yield real solution... I have no idea how to word this, please help...

Answer 1

We can see a very strange behaviour when we graph this function.
By definition, an exponential function is defined by:
\[f(x)=a^x\]
\[a \in R^+ , a \neq1\]
No, in this particular case, a=x, meaning that a will take any value x does.
But let's see what happens if I find it's derivatives:
\[y=x^x\]
\[\ln y=lnx^x \]
\[\ln y = x \ln x\]
\[\frac{ d(lny) }{ dx }=\frac{ d(xlnx) }{ dx }\]
\[\frac{ d(lny) }{ dx }=(lnx)+(\frac{ 1 }{ x })(x)\]
\[\frac{ 1 }{ y }\frac{ dy }{ dx }= 1+\ln x \]
\[\frac{ dy }{ dx }=(1+lnx)(y)\]
\[\frac{ dy }{ dx }=(1+\ln x )(x^x) \]
that sould do it, now that I have found the slope of any value of x, I see that a "Ln" appears, and that's a problem, you see, by definition, a logarithm cannot have a negative Logarithmic number.
So all the negative slopes should not be defined, just because the derivative of this function, has a logarithm in it.


Or, by definition of exponential function, it cannot have a base that is negative, because that would give me an undefined number.
See it for yourself:
https://www.google.com/search?client=opera&q=graph+x%5Ex&sourceid=opera&ie=UTF-8&oe=UTF-8#q=y%3Dx%5Ex+