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Jordan's question from Facebook (finding critical points)

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Jameson

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Jan 26, 2012
4,052
Jordan writes:

I have two questions that I've been trying to work out and I must be doing something wrong!
The first is finding all the critical points of the function:
f(x)=(6-x^2)^.5
And the second is finding all critical points of the function:
f(x)=(4-4x^2+8x)^.5

I would appreciate any input anybody could give, thanks!
 
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Jameson

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Jan 26, 2012
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Hi Jordan :)

I would rewrite the .5 as 1/2 to make it easier for calculations.

So we have \(\displaystyle f(x)=(6-x^2)^{\frac{1}{2}}\). First we need to find the derivative and remember to use the chain rule.

\(\displaystyle f'(x)=\frac{1}{2}(6-x^2)^{-\frac{1}{2}}(-2x)=\frac{-2x}{2(6-x^2)^{\frac{1}{2}}}=\frac{-x}{(6-x^2)^{\frac{1}{2}}}\)

Ok, so now we need to find where this equals 0 or where it's not defined. If the numerator equals 0 (the top) and the denominator (the bottom) does not, then the fraction equals 0 so we want to solve $-x=0$ for $x$ and it's obvious the solution is $x=0$. So we have our first critical point!

To find where the derivative is not defined we want to see when the denominator equals 0 (because dividing by 0 is bad!).

Let's look at \(\displaystyle \sqrt{6-x^2}=0\). Squaring both sides leads to \(\displaystyle 6-x^2=0\) or $x^2=6$. This has two solution, \(\displaystyle x=\sqrt{6}\) and \(\displaystyle x=-\sqrt{6}\).

To finish up, we found 3 critical points: \(\displaystyle x=0, \sqrt{6}, -\sqrt{6}\).

What do you get when you try the second one?