Question

how do you find the critical values of

f(x) = 1 + 3/x + 2/x^2 ?

Answer 1

take the derivative, set it equal to zero and solve for \(x\)

Answer 2

need help with that?

Answer 3

yes I think so.... I know that those are the steps, however I am getting confused with the 3/x and 2/x^2. and how to get the derivatives out of that.

Answer 4

the derivative of \(\frac{1}{x}\) is \(-\frac{1}{x^2}\)

Answer 5

so the derivative of \(\frac{3}{x}\) is \(-\frac{3}{x^2}\)

Answer 6

the reciprocal is an amazingly common function, so you should not "find" it you should just "know" it

Answer 7

so we do a reciprocal of 3/x^2? which is x^2 / 3?

Answer 8

as for\(\frac{2}{x^2}\) you can write it as \(2x^{-2}\) and use the almighty power rule

Answer 9

oh no
you are looking for the derivative not the reciprocal

Answer 10

or, can we do the same thing with the 3/x?

as in... 3^x-1?

Answer 11

oops, 3x^-1 *

Answer 12

\[\large \frac{d}{dx}[\frac{2}{x^2}]=\frac{d}{dx}[2x^{-2}]=-4x^{-3}=-\frac{4}{x^3}\]

Answer 13

yes you can but you shouldn't you should memorize it

Answer 14

oh... well for the time being, can the problem look like this?:

1 + 3x^-1 + 2x^-2 ?

Answer 15

then when you set it to 0 ... it would be 1 + 3x^-1 + 2x^-2 = 0?

and then find the derivatives:

So it would look like 1 + -3x^-2 + -4x^-3 ? Or am I doing it wrong?

Answer 16

I don't think I am following correctly...