# Find critical points of a function: $y = x \sqrt{4 - x^2}$

I am having trouble finding the critical points of this function, I was wondering if someone could help me out.

So I have a function $y = x\sqrt{4 - x^2}$. I know that I have to take the derivative then set that equal to 0 to find the critical points. \begin{align*} y &= x \sqrt{4 - x^2}\\ y' &= x \frac{d}{dx}\sqrt{4 - x^2} + \sqrt{4 - x^2}&& \text{(product rule)}\\ &= x\left(\frac{1}{2}(4 - x^2)^{-1/2}(-2x)\right) + \sqrt{4 - x^2} &&\text{(chain rule on square root)}\\ &= \frac{\frac{1}{2}x(-2x)}{(4-x^2)^{1/2}} + \sqrt{4-x^2}\\ &\qquad\qquad\text{(moved to denominator changed the exponent sign)} \end{align*}

This is where I am stuck, I cant seem to figure out how to proceed further, any help would be appreciated

Thanks

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Can you put them into $\LaTeX$? – Jack Oct 24 '11 at 19:48
As you said, now you set $y'=0$ and solve the resulting equation for $x$. – Ross Millikan Oct 24 '11 at 19:52
Only my second time posting here, so I don't know how to do that, thank you for editing Arturo – Hunter McMillen Oct 24 '11 at 19:52
Maybe one reason you had trouble spotting what to do is because of the two different types of notation, $\sqrt{1-4^2}$ and $(1−4x^2)^{−1/2}$, in the same formula. If you are just locating max/min, note that by symmetry if max is at $a\ge 0$, then min is at $-a$. To locate max, let's equivalently find the location of the max of the square of our function, namely $4x^2-x^4$. Here the differentiation is quick and risk-free. – André Nicolas Oct 24 '11 at 21:20

First, notice that the domain of your function is $[-2,2]$. We will only be working there.

Now, when you take the derivative, there are other critical points besides the stationary points: you must also exclude any point in the domain where the derivative is not defined. In this case, notice that if $x=-2$ or $x=2$, then the derivative is not defined, so $x=-2$ and $x=2$ are critical points.

To move further, make the expression into a single fraction: \begin{align*} \frac{\frac{1}{2}x(-2x)}{\sqrt{4-x^2}} + \sqrt{4-x^2} &= \frac{-x^2}{\sqrt{4-x^2}} + \frac{\sqrt{4-x^2}\sqrt{4-x^2}}{\sqrt{4-x^2}}\\ &= \frac{-x^2 + (4-x^2)}{\sqrt{4-x^2}}\\ &= \frac{4-2x^2}{\sqrt{4-x^2}}. \end{align*} Now, determine the points in the domain where the numerator is $0$ (we already excluded the points where the denominator is $0$).

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thank you very much this helped out a lot. – Hunter McMillen Oct 24 '11 at 20:13

Put your last expression together as one fraction by getting a common denominator. The right hand expression will be

$$\frac{4-x^2}{\sqrt{4-x^2}}$$

so the overall fraction is

$$\frac{-x^2 + 4 - x^2}{\sqrt{4-x^2}} = \frac{4 - 2x^2}{\sqrt{4-x^2}} = \frac{-2(x^2 - 2)}{\sqrt{4-x^2}}$$

which is 0 when $x = \pm \sqrt 2$.

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thanks, this helped out a lot – Hunter McMillen Oct 24 '11 at 20:13