Online domain-finding programs say that the domain of $$\frac{(x+1)^x}{x^x}$$ is x>0, and yet this function is defined for such values as x=-2, x=-4.4 and other negative values.
Why is this?
Prove that the domain of $\frac{(x+1)^x}{x^x}$ isn't simply x>0.
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functions
graphing-functions
2 Answers
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In fact we have $$f (x) =(\frac {x+1}{x})^x = (1+\frac {1}{x})^x $$
We can easily notice that $x >0$ forms one part of the domain. Also notice that for $-1 < x <0$, we get that $f (x) $ is not defined $(?)$, so on the negative sides of the real line the permissible values of $x $ is $(-\infty, -1) $.
Thus the correct domain is $$\boxed {(-\infty, -1) \cup (0,\infty)} $$
This can be confirmed by drawing the graph of the function:
Hope it helps.
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0Isn't $f(x)$ when $-1 < x < 0$ defined but complex? – 2017-02-03
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3Caution: $(x+1)^a/x^a$ is **not** necessarily the same thing as $((x+1)/x)^a$, if negative values for $x+1$ and $x$ are involved. If you don't want to allow complex numbers, the first is undefined when $x < -1$ (except for certain values of $a$). If you do allow complex numbers, you still have to be careful because the laws of exponents are not so simple. For example, $(-1)^{1/2} \cdot (-1)^{1/2} \ne ((-1) \cdot (-1))^{1/2}$. – 2017-02-03
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$f(x)$ isn't defined for x = -1 for the following reason.
$f(-1) = \frac{(-1 + 1)^{-1}}{-1^{-1}} = \frac{0^{-1}}{-1^{-1}} = \frac{\frac{1}{0}}{\frac{1}{-1}}$
For other negative values of x $$\begin{align} f(-x) &= \frac{(-x + 1)^{-x}}{-x^{-x}} \\ &= \frac{(\frac{1}{-x + 1})^x}{(\frac{1}{-x})^x} \\ &= (-x)^x(\frac{1}{1-x})^x\end{align}$$
Since $x$ is fractional you will be taking the roots of negative numbers which often results in complex numbers.