Question

Calculus question ; photo attached

Answer 1

Start by taking the derivative of your function. That will be your first step. Can you do that?

Answer 2

9+(2/x^2)=0?

Answer 3

Great. Now that you have that, solve for "x".

Answer 4

@abb0t that derivative is incorrect.

Answer 5

Hmm, is it now? Well, @Dodo1 you need to try again then.

Answer 6

how is it wrong?

Answer 7

@abb0t Good luck evaluation that term = 0 without using imaginary numbers.

@Dodo1 Forgot his - sign when taking the derivative using the power rule

Answer 8

9+(2x^(-2))=0?

Answer 9

no

Answer 10

\[9x+2x^{-1}\]\[9+(-1)2x^{-1-1}\]\[9-2x^{-2}\]\[9-2/x^2\]

Answer 11

@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

Answer 12

oh not plus ok. I see

Answer 13

next step is mutiply x^2 both side?

Answer 14

@stamp

Answer 15

@Dodo1 It is algebra I at this point. Solve for x.

Answer 16

x=Sqrt(9/2)
@stamp

Answer 17

\[9-2/x^2=0\]\[9=2/x^2\]\[x^2=2/9\]\[x=+/-(\sqrt2/3)\]

Answer 18

Ops sorry i did 9X^2+2=0

Answer 19

whats the next step?
max at _sart(2)/3?
@stamp

Answer 20

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.

Answer 21

I have to go, but this should help you enough to get started. Somebody else should be able to help you.

Answer 22

Yea, I do need to prepare derivative problems.
Are you going?

Answer 23

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.