Question

" \sqrt{ - x ^{3} } " align="absmiddle" class="latex-formula">
what would this be? The cool equation is Y=2-X^3. I'm doing Inverse Functions.

Answer 1

I'm guessing you're given the function $y(x)=2-x^3$, and you're asked to find the inverse function $y^{-1}(x)$. To do this, swap $x$ and $y$, then solve for $y$:

$x=2-y^3\implies y^3=2-x\implies y=(2-x)^{1/3}=\sqrt[3]{2-x}$

so that the inverse function is

$y^{-1}(x)=\sqrt[3]{2-x}$

Just to verify:

$y(y^{-1}(x))=y(\sqrt[3]{2-x})=2-(\sqrt[3]{2-x})^3=2-(2-x)=x$

$y^{-1}(y(x))=y^{-1}(2-x^3)=\sqrt[3]{2-(2-x^3)}=\sqrt[3]{x^3}=x$

But in case you're actually only interested in computing the square root, first we note that $\sqrt x$ (the real-valued square root) is only defined as long as $x\ge0$. So $\sqrt{-x^3}$ is defined as long as $-x^3\ge0$, or $x^3\le0$, or equivalently $x\le0$. Under this condition, we could write

$\sqrt{-x^3}=\sqrt{-x\times x^2}=\sqrt{-x}\sqrt{x^2}$

We can simplify this further, but we have to be careful. Suppose $x=-1$. Then $x^2=(-1)^2=1$. But we get the same result if $x=1$, since $x^2=1^2=1$. There are two possible values of $x$ that given the same value of $x^2$, so to capture both of them, we take $\sqrt{x^2}=|x|$, the absolute value of $x$. Then

$\sqrt{-x^3}=|x|\sqrt{-x}$

We can't simplify the square root term further than this.