# Review Question

• Dec 30th 2009, 02:27 PM
Paymemoney
Review Question
Hi
I have some question i am having trouble with:
1)If the angle between the lines which passes through the points (-2,3) and (4,0) is:
A$\displaystyle 2y=x+4$
B$\displaystyle y=\frac{-x}{2}-2$
C$\displaystyle 2y+x=4$
D$\displaystyle y=\frac{x}{2}-2$
E$\displaystyle 2y-x=4$

2)Find the derivative of $\displaystyle (4x+\frac{9}{x})^2$ and values of x at which the derivative is zero

3)Let f be differentiable for all values of x in [2,0]. The graph with equation y=f(x) has a maximum point at (1,3). The equation of the tangent at (1,3) is:
A$\displaystyle x+3y=0$
B$\displaystyle x=1$
C$\displaystyle y=3$
D$\displaystyle x-3y=0$
E$\displaystyle 3x+y=0$

P.S
• Dec 30th 2009, 02:39 PM
pickslides
Quote:

Originally Posted by Paymemoney
Hi
I have some question i am having trouble with:
1)If the angle between the lines which passes through the points (-2,3) and (4,0) is:

Have a look at this post...

http://www.mathhelpforum.com/math-he...-problems.html
• Dec 30th 2009, 02:43 PM
Paymemoney
Quote:

Originally Posted by pickslides

thanks, i know how to do this now
• Dec 30th 2009, 02:45 PM
pickslides
Quote:

Originally Posted by Paymemoney

2)Find the derivative of $\displaystyle \left(4x+\frac{9}{x}\right)^2$ and values of x at which the derivative is zero

Lets say $\displaystyle y = \left(4x+\frac{9}{x}\right)^2$

Using the chain rule

$\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

Make $\displaystyle u = 4x+\frac{9}{x}$ and $\displaystyle y = u^2$

Now find $\displaystyle \frac{dy}{du}$ and $\displaystyle \frac{du}{dx}$
• Dec 30th 2009, 03:13 PM
Paymemoney
Quote:

Originally Posted by pickslides
Lets say $\displaystyle y = \left(4x+\frac{9}{x}\right)^2$

Using the chain rule

$\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

Make $\displaystyle u = 4x+\frac{9}{x}$ and $\displaystyle y = u^2$

Now find $\displaystyle \frac{dy}{du}$ and $\displaystyle \frac{du}{dx}$

This is how far i got until i got stuck.
$\displaystyle 2(4x+\frac{9}{x})(4+\frac{9}{x^2})$

$\displaystyle 8(4x+\frac{9}{x})+\frac{9}{x^2}$

$\displaystyle 32x+\frac{72}{x}-\frac{9}{x^2}$

$\displaystyle \frac{32x^3}{x^2}+\frac{72x}{x^2}-\frac{9}{x^2}$
• Dec 30th 2009, 03:19 PM
pickslides
$\displaystyle u = 4x+\frac{9}{x} = 4x+9x^{-1}$ and $\displaystyle y = u^2$

$\displaystyle \frac{dy}{du} = 2u = 2\left(4x+\frac{9}{x}\right)$

$\displaystyle \frac{du}{dx} = 4-9x^{-2}$

$\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

$\displaystyle \frac{dy}{dx} = 2\left(4x+\frac{9}{x}\right)\times \left(4-9x^{-2}\right)$

Now expnad using the F.O.I.L method. Do you know this method?
• Dec 30th 2009, 03:38 PM
Paymemoney
Quote:

Originally Posted by pickslides
$\displaystyle u = 4x+\frac{9}{x} = 4x+9x^{-1}$ and $\displaystyle y = u^2$

$\displaystyle \frac{dy}{du} = 2u = 2\left(4x+\frac{9}{x}\right)$

$\displaystyle \frac{du}{dx} = 4-9x^{-2}$

$\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

$\displaystyle \frac{dy}{dx} = 2\left(4x+\frac{9}{x}\right)\times \left(4-9x^{-2}\right)$

Now expnad using the F.O.I.L method. Do you know this method?

oh ok yeh i know what the method is now.