Answer
(a)$g^{-1}=\sqrt{\frac{4625-r}{18500}}$
It represents distance from central axis ($g^{-1}$) for given velocity ($r$).
(b) $g^{-1}(30)\approx 0.4984$
It means, that for $30$ $cm/s$ (blood speed) its distance from central axis is $\approx0.4984cm$
Work Step by Step
$g(r)=18500(0.25-r^2)$
(a) To find $g^{-1}$ we will go through the following steps:
First write the function in terms of $y$ and $r$ (Originally $x$):
$y=18500(0.25-r^2)$
Now replace $y$ by $r$ and vice versa:
$r=18500(0.25-y^2)$
And solve it for $y$:
$r=4625-18500y^2$
$18500y^2=4625-r$
$y=\sqrt{\frac{4625-r}{18500}}$
$g^{-1}(r)=\sqrt{\frac{4625-r}{18500}}$
It represents distance (the radius) from central axis for given velocity.
(b) $g^{-1}(30)=\sqrt{\frac{4625-30}{18500}}\approx\sqrt{0.2484}\approx0.4984$
It means, that for $30$ $cm/s$ (blood speed) its distance from central axis is $\approx0.4984cm$