Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

Calculus question ; photo attached

Calculus1
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

1 Attachment
Start by taking the derivative of your function. That will be your first step. Can you do that?
9+(2/x^2)=0?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Great. Now that you have that, solve for "x".
@abb0t that derivative is incorrect.
Hmm, is it now? Well, @Dodo1 you need to try again then.
how is it wrong?
@abb0t Good luck evaluation that term = 0 without using imaginary numbers. @Dodo1 Forgot his - sign when taking the derivative using the power rule
9+(2x^(-2))=0?
no
\[9x+2x^{-1}\]\[9+(-1)2x^{-1-1}\]\[9-2x^{-2}\]\[9-2/x^2\]
@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
oh not plus ok. I see
next step is mutiply x^2 both side?
@Dodo1 It is algebra I at this point. Solve for x.
x=Sqrt(9/2) @stamp
\[9-2/x^2=0\]\[9=2/x^2\]\[x^2=2/9\]\[x=+/-(\sqrt2/3)\]
Ops sorry i did 9X^2+2=0
whats the next step? max at _sart(2)/3? @stamp
You 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.
I have to go, but this should help you enough to get started. Somebody else should be able to help you.
Yea, I do need to prepare derivative problems. Are you going?
f(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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question