# More function problems...

• Feb 9th 2010, 05:37 PM
Jools
More function problems...
Find the points on the curve y = 2/3x + 2 where the tangent is parallel to the line y = (-3/2x) - 1

I know the slope of the tangent is -3/2 as it is parallel to the line y = (-3/2x) - 1.

Therefore m = -3/2

So I can express the curve as...

y = 2(3x - 2)^-1

Using a combination of the product rule and chain rule I have come up with:

= -6/(3x - 2)^2 + 1/(3x - 2)

So...

-3/2 = -6/(3x - 2)^2 + 1/(3x - 2)

But that's where I get hung up. Can I express that as the following then solve for x?

-3/2 = -5/(3x - 2)^2 + (3x - 2)

Thanks!
• Feb 9th 2010, 06:44 PM
Soroban
Hello, Jools!

Your derivative is wrong . . .

Quote:

Find the points on the curve $y \,=\, \frac{2}{3x + 2}$
where the tangent is parallel to the line $y \,=\, -\tfrac{3}{2}x - 1$

I know the slope of the tangent is $-\tfrac{3}{2}$.

I can express the curve as: . $y \:=\:2(3x - 2)^{-1}$

The derivative is: . $\frac{dy}{dx} \:=\:-2(3x+2)^{-2}\cdot3 \:=\:\frac{-6}{(3x+2)^2}$

So we have: . $\frac{-6}{(3x+2)^2} \;=\;-\frac{3}{2} \quad\Rightarrow\quad -12 \;=\;-3(3x-2)^2$

. . $(3x+2)^2 \;=\;4 \quad\Rightarrow\quad 3x + 2 \;=\;\pm2 \quad\Rightarrow\quad 3x \;=\;-2\pm2$

Therefore: . $x \;=\;\frac{-2\pm2}{3} \;=\;\begin{Bmatrix}\dfrac{-2+2}{3} &=& 0 \\ \\[-3mm] \dfrac{-2-2}{3} &=& \text{-}\dfrac{4}{3} \end{Bmatrix}$

• Feb 9th 2010, 06:49 PM
Lord Voldemort
Oh my, it seems to me like you're complicating it.
The way I see it, you're looking at a spot on the derivative of y where y' = -3/2.
$y = \frac{2}{3x}+2$
$y' = \frac{d}{dx}\frac{2}{3x}+\frac{d}{dx}(2)$
Now, the derivative of a constant is 0, so the 2nd term is 0 now.
Also, remember that 2/3 is also a constant and can thus be taken out of the d/dx.
$y' = -\frac{2}{3x^2}$
$-\frac{3}{2} = -\frac{2}{3x^2}$
$9x^2 = 4$
I trust you can solve this? (I don't know how to do the plus or minus thing with LaTex)
• Feb 9th 2010, 07:03 PM
Jools
First I need to point out an error I made here... The curve in question should actually be y = 2/3x - 2, not y = 2/3x + 2. Sorry... my bad. Also, I'm not familiar with dy/dx... How is this being applied, and what does each variable represent? Thanks!
• Feb 9th 2010, 07:43 PM
Lord Voldemort
dy/dx means taking the derivative of y with respect to x, or differentiating y with x as the variable of differentiation.
But remember, though y = 2/3x - 2, it is still y = 2(3x-2)^-1
y' = -6(3x-2)^-2
Then plug in y = -3/2
(3x-2)^2 = 4
3x - 2 = 2 or 3x - 2 = -2
x = 4/3 or x = 0
• Feb 9th 2010, 07:54 PM
Jools
So -6(3x - 2)^-2 I understand, but when we use the product rule, it should equate to F ' (x) = f(x)g ' (x) + f ' (x)g(x) shouldn't it? So what is happening with the second half of the formula?...
• Feb 10th 2010, 02:24 AM
HallsofIvy
And please use parentheses to clarify what you mean!

You talk about the line "y = (-3/2x) - 1" so that is clearly supposed to be (-3/2)x but you also have "y = 2/3x + 2" which we clearly are supposed to interpret as y= 2/(3x)+ 2 because if we interpret it the same way, as y= (2/3)x+ 2, is a line with slope 2/3 and the two lines are perpendicular.
• Feb 10th 2010, 04:33 AM
Jools
My apologies. To clarify.... it it 2 / (3x - 2)