# How do you graph f(x) = xsqrt(4-x^2)

Feb 20, 2015

First you need to be sure to avoid $x$ values that makes the argument of your square root negative (you cannot find a real solution of a negative square root). So you set:
$4 - {x}^{2} \ge 0$ and:
$\left(2 - x\right) \left(2 + x\right) \ge 0$
so:
$x \ge - 2$
$x \le 2$
and $- 2 \le x \le 2$
Outside this interval your function does not exist.

So, now we choose values for $x$ and evaluate the correspondent $y$:
$x = 2 \implies y = 0$
$x = - 2 \implies y = 0$
$x = 0 \implies y = 0$
We have a graph that passes through $y = 0$ 3 times; probably our function changes quadrant in doing so.

For $x$ approaching $- 2$ our function gives negative values (for example when $x = - 1.9 \implies y = - 1.2$) and when approaching $2$ it has positive values (for example when $x = 1.9 \implies y = 1.2$). The graph of our function starts from $x = - 2$ going downwards, rises passing through $x = 0$ and then bends again to end up in $x = 2$. I must have a minimum and a maximum somewhere.

I evaluate the derivative of the function and set it equal to zero to find these points:
$f ' \left(x\right) = \sqrt{4 - {x}^{2}} - {x}^{2} / \left(\sqrt{4 - {x}^{2}}\right)$

setting it equal to zero and manipulating it I got:
$4 - 2 {x}^{2} = 0$ which gives $x = \pm \sqrt{2}$
so you get:
$x = - \sqrt{2} \implies y = - 2$ minimum and
$x = \sqrt{2} \implies y = 2$ maximum

Finally: 