• Nov 25th 2006, 07:07 AM
jmailloux
I have this assignment due wednesday.. and I've started it.. but I want to make sure I'm on the right track.. and some of them I have no clue where to start. Anyways.. if anyone can help it would be much appreciated

1. Determine the derivative of
f(x)= 2x^2 - 1
.............x

2. The graph of
f(x)= ax + b
......(x -1)(x-4)
has a horizontal tangent line at (2,-1). Determine the values of a and b.

3. The tangent line to the curve defined by
g(x)= x^4 - 2x^3 +3
..............-6x
at x= -1 intersects the curve at two other points P and Q. Determine the coordinates of P and Q algebraically. Illustrate this situation with a graph. (to graph g you need to create a table of values)
HINT: You need to research L'hopital's rule in order to complete this question

I have no clue what L'hopital's rule is, because we've never learned it, so if someone could also explain that to me.. that would be great.

Thank-you!
• Nov 25th 2006, 07:57 AM
earboth
Quote:

Originally Posted by jmailloux
.. if anyone can help it would be much appreciated

1. Determine the derivative of
f(x)= 2x^2 - 1
.............x

...

Hello,

$f(x)=\frac{2x^2-1}{x}=2x-\frac{1}{x}$. Now calculate the derivative:
$f'(x)=2+\frac{1}{x^2}$. Using quotient rule:
$f'(x)=\frac{x\cdot (2\cdot 2x)-(2x^2-1)\cdot 1}{x^2}=\frac{4x^2-2x^2+1}{x^2}=\frac{2x^2+1}{x^2}=2+\frac{1}{x^2}$

EB
• Nov 25th 2006, 08:21 AM
earboth
Quote:

Originally Posted by jmailloux
.. if anyone can help it would be much appreciated

2. The graph of
f(x)= ax + b
......(x -1)(x-4)
has a horizontal tangent line at (2,-1). Determine the values of a and b.
...

Hello,

you have the function:
$f(x)=\frac{ax+b}{(x-1)(x-4)}$. And you know f(2) = -1 and f'(2) = 0:

$f(2)=\frac{2a+b}{(2-1)(2-4)}=-1\ \Longleftrightarrow\ 2a+b=2$

$f'(x)=\frac{(x-1)(x-4)\cdot a-(ax+b)\cdot (2x-5)}{((x-1)(x-4))^2}$. Don't expand that. Plug in x = 2. Then f'(2) = 0. Multiply by the denominator :

$-2a+2a+b = 0$. Therefore b = 0 and a = 1.

$f(x)=\frac{x}{(x-1)(x-4)}$

EB
• Nov 26th 2006, 01:48 AM
earboth
Quote:

Originally Posted by jmailloux
... if anyone can help it would be much appreciated
...
3. The tangent line to the curve defined by
g(x)= x^4 - 2x^3 +3
..............-6x
at x= -1 intersects the curve at two other points P and Q. Determine the coordinates of P and Q algebraically. Illustrate this situation with a graph. (to graph g you need to create a table of values)
HINT: You need to research L'hopital's rule in order to complete this question

I have no clue what L'hopital's rule is, because we've never learned it, so if someone could also explain that to me.. that would be great.
Thank-you!

Hello,

I'll show you a way where you don't have to use rule de l'Hopital. (There is actually no need to use this rule!)

I'll use: $g(x)=\frac{x^4-2x^3+3}{-6x}=-\frac{1}{6}x^3+\frac{1}{3}x^2-\frac{1}{2x},\ x\neq0$

Then g(-1) = 1

$g'(x)=-\frac{1}{2}x^2+\frac{2}{3}x+\frac{1}{2x^2}$
Thus $g'(-1)=-\frac{2}{3}$

The tangent t becomes: $y= t(x)=-\frac{2}{3}x+\frac{1}{3}$

Calculating the intercept mean: g(x) = t(x):
$-\frac{1}{6}x^3+\frac{1}{3}x^2-\frac{1}{2x}=-\frac{2}{3}x+\frac{1}{3}$
Multiply by (-6x):
$x^4-2x^3-4x^2+2x+3=0$. You already know that x = -1 must be a solution of this equation. So divide the LHS of the equation by (x+1) and you'll get:
$x^3-3x^2-x+3=0$
$x^2(x-3)-(x-3)=0$
$(x^2-1)(x-3)=(x+1)(x-1)(x-3)0$.

So you get the additional solutions: x = 1 and x = 3.

I'll leave the rest for you. Good luck!

EB
• Nov 26th 2006, 04:42 AM
earboth
Quote:

Originally Posted by jmailloux
... if anyone can help it would be much appreciated
...
3. The tangent line to the curve defined by
g(x)= x^4 - 2x^3 +3
..............-6x
...

Hello again,

for confirmation only I send you the graph of the function and the tangent line.

EB