Quick Question

[doughishere]

\(\displaystyle g(x)= \frac{\sqrt[3]{x}}{x^2+1}\)

The domain of this is all real numbers because even if x is equal to 0 then the denominator still equals one right? Even if write it in the alternate form bring the cube root to the denominator.

 

[Dr.Peterson]

\(\displaystyle g(x)= \frac{\sqrt[3]{x}}{x^2+1}\)

The domain of this is all real numbers because even if x is equal to 0 then the denominator still equals one right? Even if write it in the alternate form bring the cube root to the denominator.

Your conclusion is correct. The way I would explain it is that the denominator is always at least 1, so there is never a division by zero; and the cube root (unlike the square root) is defined for all real numbers.

 

[JeffM]

\(\displaystyle g(x)= \frac{\sqrt[3]{x}}{x^2+1}\)

The domain of this is all real numbers because even if x is equal to 0 then the denominator still equals one right? Even if write it in the alternate form bring the cube root to the denominator.

Yes, the domain of g(x) as defined is all real numbers, but your reasoning is not quite right. We are not concerned with where x equals zero. We are concerned with where the denominator MAY be zero. And there is no real value of x such that the denominators is zero. Got that?

\(\displaystyle h(x) = \dfrac{x}{(x^2 + 1)\sqrt[3]{x^2}} = g(x) \text { if and only if } x \ne 0.\)

Thus, you cannot say that g(x) = h(x). The domain of h(x) is not the same as the domain of g(x), and so the functions are not identical. The function h(x) is not defined at x = 0 whereas g(x) is defined for all real values of x.

 

[mmm4444bot]

Quick Qustion …

Even if write it in the alternate form bring the cube root to the denominator.

Seems like you're trying to make a conditional statement, but some of the words are missing.

What is "the" alternate form? Maybe, you're thinking about exponential notation.

\(\displaystyle \dfrac{1}{(x^2+1) \cdot x^{-1/3}}\)

Or, maybe the following.

\(\displaystyle \dfrac{1}{\dfrac{x^2+1}{\sqrt[3]{x}}}\)

As Jeff noted, you need to be mindful of the domains, when using alternate forms.