The domain of $\left(\frac{x^2-3}{x^2+4}\right)^{2x^3}$

algebra-precalculusfunctions

I thought the domain of the function
$$f(x)=\left(\frac{x^2-3}{x^2+4}\right)^{2x^3}$$
is $[\sqrt{3},\infty)$, by using the formula
$$x^y = \exp\bigl(y\ln(x)\bigr).$$
But then why I can compute
$$
f(1) = \frac{4}{25} \text{ or } f(0) = 1?
$$
Is this valid? Any help would be appreciated. Thanks. 🙂

Best Answer

If the exponent in this function had a fixed sign, there would be only one portion of this specific function that would determine the domain of it, which would be the denominator. But this denominator being $x^2+4$ is always positive for $x\in \Bbb R$ so that part is fine.

But this function has an exponent that can be negative, i.e., $2x^3\in(-\infty,\infty)$. So then we have to check the inverted fraction's denominator for invalid inputs, and we find that $x=-\sqrt 3$ is not part of this function's domain.

Edit for negative to exponents: since the numerator can also be negative with a possible non-rational exponent, these values should be excluded...

Edit: the effective domain of this function is $(-\infty,-\sqrt 3)\cup[\sqrt 3,\infty)\cup\{a\in(-\sqrt 3,\sqrt 3):(-1)^{2a^3}\in\Bbb R\}.$

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