more difficult questions (for me)?

Question

1) f(X)=x^4-3x^2+2
f'(x)= 4x^3-6x
line tangent to the graph of f(x) is y=6x-8
Find the x-coordinate of each point at which the line tangent to the graph of f is parallel to the line y=-2x+4.


2)f(x)= ln(X^2-9)

QUESTION: Determine the symmetry of the graph of f.
Question: Write a formula for the inverse of f, for x>3.



3. f(x)= x^3-x/x^3-4x

QUESTION: Write an equation for each vertical and each horizontal asymptote to the graph f. (Show all work, including limits to infinty to get the answers)

QUESTION: Describe the symmetry of the graph of f. Show all work using your symmetry test.

Answer 1

For #1,
f'(x)= 4x^3-6x
You need to find the x values where the slope is 2, or f'(x)=2
4x^3-6x=2
4x^3-6x-2=0
Solving numerically, x=-1, -0.366, 1.366 (the last two are approximations, not exact values)

#2
y=ln(x^2-9)
Since you can't take the log(0), x^2-9>0
(x-3)(x+3)>0
x=+/-3
But since it's an inequality, you have to test to find out where the values hold truth. Checking, x<-3 or x>3 give us truth values. Therefore, you can argue that the function will be symmetric about the y axis.

For the inverse, switch variables then solve for y:
x=ln(y^2-9)
e^x=y^2-9
e^x + 9 = y^2
y=+/-sqrt(e^x + 9)

#3
f(x)= (x^3-x)/(x^3-4x)

To find vertical asymptotes, set the denominator = 0
x^3-4x=0
x(x^2-4)=0
x(x-2)(x+2)=0
x=0,2,-2

To find horizontal asymptotes, take the limit of the function as x->inf
Lim x->inf of x^3-x/x^3-4x
Divide through by the highest power of x:
Lim x->inf of 1-(1/x^2)/(1-(4/x^2))
As x->inf, the terms that are divided will go to zero, leaving
y=1 as the horizontal asymptote.

I can't help you with the graph much, other than stating that the asymptotes cut the xy graph into 6 sections. All you need to do is pick an x value in each of the 3 intervals between vertical asymptotes and determine whether the corresponding y value is below or above the horizontal asymptote. Once you know which of the 3 sextants that the graph is in, you can fill in the whole graph since it will merely "skirt" each asymptote line out to the appropriate infinity.

Viola!

Answer 2

4x^3-6x=-2
4x^3-6x+2=0 x=1 is a solution
so
(x-1)(4x^2+4x-2)=0
2x^2+2x-1=0 x=((-2+-sqrt(12)/4 = 1/2(-1+-sqrt3)
2)if you change x into -x the y value remains so 0y is an axis of symmetry
e^y= x^2-9 so x= sqrt(e^y+9) for x >3
Interchange x and y so y=sqrt(e^x+9)
3)(x^3-x)/(x^3-4x) is for all x not =0 ( x^2-1)/(x^2-4)
Vertical asymptotes x=+-2
Horizontal asymptote y=1
To question 1
if y=6x+8 is a tangent
4x^3-6x-6=0 which has a root between 1 and 2 x=1.56746837485