Question

Find the relative extrema of y=4x+x^-1

Answer 1

What is extrema ? Maxima or Minima

Answer 2

maxima

Answer 3

Ah then Here you go
\[\frac{dy}{dx} = 4 - \frac{1}{x^2}\]
\[Put\:\:\:\:\:\:\frac{dy}{dx} = 0\]
\[4 = \frac{1}{x^2}\]
\[x^2 = \frac{1}{4}\]
\[x = \pm \

Answer 4

\[x = \pm \frac {1}{2}\]

Answer 5

Sorry \[\frac{-1}{2}\]is the maxima

Answer 6

It is correct WooBin do you wan to ask anything

Answer 7

umm...i think its better that you put some explanation so i can fully understand your asnwer. but this okay it helps a lot.

Answer 8

Sure

Answer 9

Now when we take the second derivative

\[\frac{d^2y}{dx^2} = \frac{2}{x^3}\]

Answer 10

Then we put the points -1/2 , +1/2

Answer 11

\[x= \frac{-1}{2}\]
\[-16\]------------------->So this is Maxima

\[x=+\frac{1}{2}\]
\[16\]-------------------->This is Minima

Answer 12

Now we have -1/2 has maxima point and 1/2 as minima point

Answer 13

Now woobin do you understand it now ?

Answer 14

Call me whenever you have some doubts :D

Answer 15

A visual aid is attached.