Critical Points

**spyder12** · February 12th 2013, 05:40 PM

What are the critical point(s) of h(x)=(9-4x^2)^(1/2) ?

**Prove It** · February 12th 2013, 05:46 PM

The critical points are where $\displaystyle \begin{align*} h'(x) = 0 \end{align*}$ .

**spyder12** · February 12th 2013, 05:49 PM

after setting the derivative to 0 I got x=0 is that correct?

**Prove It** · February 12th 2013, 06:01 PM

What did you get for the derivative?

**spyder12** · February 12th 2013, 06:03 PM

I used my calculator for that I got ( -4x ) / ( -4 ( x^2 - 2.25) ^ (1/2) )

**Prove It** · February 12th 2013, 10:54 PM

Well if you were going to do it by hand, do it this way...

$\displaystyle \begin{align*} h &= \left( 9 - 4x^2 \right)^{\frac{1}{2}} \end{align*}$

Let $\displaystyle \begin{align*} u = 9 - 4x^2 \end{align*}$ so that $\displaystyle \begin{align*} h = u^{\frac{1}{2}} \end{align*}$ .

Then $\displaystyle \begin{align*} \frac{du}{dx} = -8x \end{align*}$ and $\displaystyle \begin{align*} \frac{dh}{du} = \frac{1}{2}u^{-\frac{1}{2}} = \frac{1}{2\left( 9 - 4x^2 \right) ^{\frac{1}{2}}} \end{align*}$

So $\displaystyle \begin{align*} \frac{dh}{dx} = \frac{-8x}{2 \left( 9 - 4x^2 \right)^{\frac{1}{2}} } = \frac{-4x}{\left( 9 - 4x^2 \right)^{\frac{1}{2}}} \end{align*}$

Yes, you are correct that when $\displaystyle \begin{align*} \frac{dh}{dx} = 0 \end{align*}$ then $\displaystyle \begin{align*} x = 0 \end{align*}$ . What is the value of h when x = 0?

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