# Critical Points

• Feb 12th 2013, 05:40 PM
spyder12
Critical Points
What are the critical point(s) of h(x)=(9-4x^2)^(1/2) ?
• Feb 12th 2013, 05:46 PM
Prove It
Re: Critical Points
The critical points are where \displaystyle \begin{align*} h'(x) = 0 \end{align*}.
• Feb 12th 2013, 05:49 PM
spyder12
Re: Critical Points
after setting the derivative to 0 I got x=0 is that correct?
• Feb 12th 2013, 06:01 PM
Prove It
Re: Critical Points
What did you get for the derivative?
• Feb 12th 2013, 06:03 PM
spyder12
Re: Critical Points
I used my calculator for that I got ( -4x ) / ( -4 ( x^2 - 2.25) ^ (1/2) )
• Feb 12th 2013, 10:54 PM
Prove It
Re: Critical Points
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?