1. ## Differentiation Problem

Hi

I have a question on a paper that I have to solve though I have become unsure as to what to do.

The problem is as follows

Given that y=16x+x^-1, find the two values of x for which dy/dx = 0.

When I differentiate I get 16 + x^-2 or 16 + 1/x^2

to remove the fraction I multiply both sides by x^2 giving

16x^2 + 1 = 0

so dy/dx = 0 -> 16x^2 + 1 = 0.

I thought to obtain the two values of x you have to factorise the quadratic, and I am not sure how to do this given the form of the derived equation.

Any help with this would be gratefully appreciated.

Thank You

David

2. Originally Posted by DavidRUK
Hi

I have a question on a paper that I have to solve though I have become unsure as to what to do.

The problem is as follows

Given that y=16x+x^-1, find the two values of x for which dy/dx = 0.

When I differentiate I get 16 + x^-2 or 16 + 1/x^2 Mr F says: This is wrong. It should be 16 - x^-2 or 16 - 1/x^2.

to remove the fraction I multiply both sides by x^2 giving

16x^2 + 1 = 0

so dy/dx = 0 -> 16x^2 + 1 = 0.

I thought to obtain the two values of x you have to factorise the quadratic, and I am not sure how to do this given the form of the derived equation.

Any help with this would be gratefully appreciated.

Thank You

David
..

3. Originally Posted by DavidRUK

Given that y=16x+x^-1, find the two values of x for which dy/dx = 0.

When I differentiate I get 16 + x^-2 or 16 + 1/x^2

This is not completly correct

$
y=16x+x^{-1} \implies \frac{dy}{dx}= 16-x^{-2}
$

Now finish it off as you were.

4. Other than that error in differentiating it looks like your main doubt was:
I thought to obtain the two values of x you have to factorise the quadratic, and I am not sure how to do this given the form of the derived equation.
Since there is no x^1 term in the quadratic, you don't need to factorise anything.

You have $\frac{dy}{dx}= 16- \frac{1}{x^2} = 0$

I wont tell you the full solution unless you want me too; I'm sure you can see what to do. You need to do some simple algebra and then remember to think about whether you need +.

5. Sorry about that. I got so caught up in the problem i forgot to bring the -1 down to change the sign.

I am still not sure how to find two values for x though.

6. Originally Posted by DavidRUK
Sorry about that. I got so caught up in the problem i forgot to bring the -1 down to change the sign.

I am still not sure how to find two values for x though.
I don't see how you can be studying calculus and not know how to solve an equation like $16- \frac{1}{x^2} = 0$. That is very worrying and suggests life is going to be very tough for you unless you get remedial help in pre-calculus ....

The most obvious approach is

$\frac{1}{x^2} = 16$

$\Rightarrow x^2 = \frac{1}{16}$

etc.

7. You have no idea...
Try doing anything with a screaming baby and a sick wife..
Also I have been up since 4:30AM.

Not good.

Thanks though.. I will go and practice.

8. Originally Posted by DavidRUK
You have no idea...
Try doing anything with a screaming baby and a sick wife..
Also I have been up since 4:30AM.

Not good.

Thanks though.. I will go and practice.
I'm sorry to hear that.

(Note that 99 times out of 100 there is no "screaming baby and a sick wife" ....)