What are the critical point(s) of h(x)=(9-4x^2)^(1/2) ?
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The critical points are where .
after setting the derivative to 0 I got x=0 is that correct?
What did you get for the derivative?
I used my calculator for that I got ( -4x ) / ( -4 ( x^2 - 2.25) ^ (1/2) )
Well if you were going to do it by hand, do it this way... Let so that . Then and So Yes, you are correct that when then . What is the value of h when x = 0?
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