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!

2. 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

3. 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

4. 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

5. 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