find the negative value of x where the tangent line to the graph of f(x)=3*x+(1/x) has a slope=-2?

Question

zepdrix · Answer

So we want to find an x value for which f'(x)=-2.

They said `find the negative value of x`, which implies we'll probably find more than one x value when we get to our solution.

zepdrix · Answer

Try taking the derivative, what do you get? :)
f'(x)=?

anonymous · Answer

well 3*x is just 3, and 1/x is -1/x^2

zepdrix · Answer

$$\Large f'(x)\quad=\quad 3-\frac{1}{x^2}$$
Ok good, so we want to find a location where the slope of the tangent line is -2.$$\Large -2\quad=\quad3-\frac{1}{x^2}$$

anonymous · Answer

ok i tried this earlier but i must have just done something wrong. so do you just solve for x?

zepdrix · Answer

Yes, solve for x.
See how we have a square on the x?
We'll end up getting 2 solutions, positive and negative.
We'll want to take the negative one as our final answer.

anonymous · Answer

so can't you bring the x^2 into the numerator just by making it negative?

zepdrix · Answer

There are a bunch of different ways to solve for x, just depends what you are most comfortable with.
Personally I would multiply both sides by x^2 giving us,$$\Large -2x^2\quad=\quad 3x^2-1$$
But that's cause I don't like dealing with those ugly fractions.

zepdrix · Answer

Yes you could change x^2 to x^{-2} and bring it to the numerator, but I don't think that's the route you wanna take :p

anonymous · Answer

so how did you get the x^2 on the -2 as well?

zepdrix · Answer

Multiplying both sides by x^2 gives us,$$\Large x^2\left(-2\right) \quad=\quad x^2\left(3-\frac{1}{x^2}\right)$$
Which simplifies to,$$\Large -2x^2\quad=\quad 3x^2-1$$

anonymous · Answer

sorry it took me so long to respond, the website stopped working

zepdrix · Answer

yes it did :( grumble grumble..

anonymous · Answer

lol

anonymous · Answer

so can we combine -2*x^2 and 3*x^2?

zepdrix · Answer

mhm

zepdrix · Answer

$$\Large 0=5x^2-1$$

anonymous · Answer

so its + or - 1/5?

zepdrix · Answer

Woops! Don't forget to take the square root! :O
Maybe just a typo.
$$\Large x\quad=\quad\pm \frac{1}{\sqrt5}\quad=\quad\pm\frac{\sqrt5}{5}$$

zepdrix · Answer

We should probably write it like that  ^ to be legit.

zepdrix · Answer

So which value of x do we want?
The positive or negative? +_+

anonymous · Answer

ok i got it and checked it and its all good!!! the - one of course!

zepdrix · Answer

yay team \c:/

anonymous · Answer

yay! thank you!!! now i can sleep haha