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anonymous
 3 years ago
Calculus question ; photo attached
anonymous
 3 years ago
Calculus question ; photo attached

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abb0t
 3 years ago
Best ResponseYou've already chosen the best response.1Start by taking the derivative of your function. That will be your first step. Can you do that?

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.1Great. Now that you have that, solve for "x".

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1@abb0t that derivative is incorrect.

abb0t
 3 years ago
Best ResponseYou've already chosen the best response.1Hmm, is it now? Well, @Dodo1 you need to try again then.

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1@abb0t Good luck evaluation that term = 0 without using imaginary numbers. @Dodo1 Forgot his  sign when taking the derivative using the power rule

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1\[9x+2x^{1}\]\[9+(1)2x^{11}\]\[92x^{2}\]\[92/x^2\]

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1@Dodo1 How did you plan on solving 9 + 2/x^2 = 0? It is not possible using real numbers because any real number squared is positive, this was the first clue that something was wrong with the derivative

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh not plus ok. I see

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0next step is mutiply x^2 both side?

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1@Dodo1 It is algebra I at this point. Solve for x.

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1\[92/x^2=0\]\[9=2/x^2\]\[x^2=2/9\]\[x=+/(\sqrt2/3)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ops sorry i did 9X^2+2=0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0whats the next step? max at _sart(2)/3? @stamp

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1You do not need to tag me in every response, Dodo, I am here. These values, x = +/ c, are critical points. They are max or mins, we do not know until we consider each point with respect to the second derivative. You have been familiarized with the second derivative test in lecture, now we are using it to see which of these critical points are max, mins, etc.

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1I have to go, but this should help you enough to get started. Somebody else should be able to help you.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Yea, I do need to prepare derivative problems. Are you going?

stamp
 3 years ago
Best ResponseYou've already chosen the best response.1f(x) = http://www.wolframalpha.com/input/?i=9x%2B2%2Fx f'(x) = http://www.wolframalpha.com/input/?i=derivative+of+9x%2B2%2Fx f''(x) = http://www.wolframalpha.com/input/?i=second+derivative+of+9x%2B2%2Fx critical points where f'(x) is 0 = http://www.wolframalpha.com/input/?i=solve+derivative+of+9x%2B2%2Fx graph of f(x) http://www.wolframalpha.com/input/?i=plot+9x%2B2%2Fx Visually one can observe that the negative critical point is a maximum, while the positive critical point is a minimum. Analytically, you can observe the change of f''(x) to the left of the critical point and to the right of the critical point. Since f'(  c ) goes from + to  ,  c is a maximum. Since f'( + c ) goes from  to +, + c is a minimum.
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