1. ## analyze function

- F(x)= x-5/x^2-2x-15
Part A) Find the domain of f(x)
Part B) Find the x and y intercepts for f(x)
Part C) find the vertical and horizontal asymtotes for f(x)
Part D) find the derivitive for f(X) and the critical points
Part E) find the second derivative and the inflection points of F(X)

Find the area bound betweeen the curves f(x)=x^2-4x+3 and f(x)= -2x+3

Find the volume of the solid formed by revolving the region bounded by f(x)= 4x-x^2 and g(x)=2 around the line y=2

Find the volume of the solid formed by revolving the region bounded by f(x)=4x-x^2 and g(x)=2 around the line x=5

The radius r of a sphere is increasing at a rate of .7 inches per minute. Find the rate of change of the volume when the radius is 4 inches. V=4/3(pieR^3)

2. Originally Posted by vc15ao4
- F(x)= x-5/x^2-2x-15
Part A) Find the domain of f(x)
Part B) Find the x and y intercepts for f(x)
Part C) find the vertical and horizontal asymtotes for f(x)
Part D) find the derivitive for f(X) and the critical points
Part E) find the second derivative and the inflection points of F(X)
Here's the first question. read http://www.mathhelpforum.com/math-he...-help-plz.html for info on vertical and horizontal asymptotes.

A)Find the domain of f(x)
the domain of a function is the set of all inputs (in this case x-values) for which the function is defined. it is almost always easier to find the x's for which the function is NOT defined, and say it's defined for all x's but those. so let's do that.

this is a rational function, it is defined everywhere as long as it's denominator is not zero.

so we need to find where x^2-2x-15 is zero.

x^2-2x-15 = 0
=> (x - 5)(x + 3) = 0
=> x = 5 and x = -3

these are the x's that make the denominator zero, so the domain is given by:

dom(f) = {x : x not= 5 and x not= -3}

or in interval form:

dom(f) = (-infinity, -3)U(-3,5)U(5, infinity)

B) Find the x and y intercepts for f(x)

for the x-intercepts, set y = 0
=> 0 = (x-5)/(x^2-2x-15)
=> 0 = x - 5 ...........since x^2-2x-15 can't be zero
=> x = 5 ..........x-intercept

for the y-intercept, set x = 0
=> y = (0-5)/[0^2-2(0)-15] = -5/-15 = 1/3 ........y-intercept

C) find the vertical and horizontal asymtotes for f(x)

vertical asymptotes occur at x-values for which the function is undefined, so it will be all x's NOT in the domain. we did the work in the first part so i won't do it again.

our vertical asymptotes are: x = 5 and x = -3

for the horizontal asymptote we have y = 0, since the degree of the numerator is smaller than the degree of the denominator. the official way to do this question is to take the limit as x goes to infinity and -infinity, if the result is a number, that's the horizontal asymptote

D) find the derivitive for f(X) and the critical points

ok, this is a rational function, let's use the quotient rule to find the derivative.

Recall, Quotient rule:

(u/v)' = (v*u' - u*v')/v^2

f(x) = (x-5)/(x^2-2x-15)
By the Quotient rule:
=> f ' (x) = [(x^2 - 2x - 15) - (x - 5)(2x - 2)]/[(x^2-2x-15)^2]
.............= (x^2 - 3x -10)/[(x^2-2x-15)^2]
...we can factorize and simplify this a bit, but i won't bother

critical points occur when the derivative is zero.

for critical points, set f ' (x) = 0
=> (x^2 - 3x -10)/[(x^2-2x-15)^2] = 0
=> x^2 - 3x - 10 = 0
=> (x - 5)(x + 2) = 0
=> x = 5, x = -2
x = 5 is an asymptote, so x = -2 is the only critical point. (actually, the (x - 5) would cancel if we simplified f ' (x), so don't worry about it too much)

E) find the second derivative and the inflection points of F(X)

for the second derivative, we again use the quotient rule, but this time on the first derivative.

f '' (x) = (f ' (x))' = [(x^2 - 3x -10)/[(x^2-2x-15)^2]]'
.......................= {[(x^2-2x-15)^2]*(2x - 3) - (x^2 - 3x - 10)(2*(x^2 - 2x - 15)*(2x - 2)}/[[(x^2-2x-15)^4]

now i really don't want to simplify that!

for the POSSIBLE inflection points, set f '' (x) = 0

=> {[(x^2-2x-15)^2]*(2x - 3) - (x^2 - 3x - 10)(2*(x^2 - 2x - 15)*(2x - 2)}/[[(x^2-2x-15)^4] = 0
=> [(x^2-2x-15)^2]*(2x - 3) - (x^2 - 3x - 10)(2*(x^2 - 2x - 15)*(2x - 2) = 0
=> (x - 5)^2*(x + 3)^2*(2x - 3) - 2(x - 5)(x + 2)(x - 5)(x + 3)(2x - 2) = 0
=> [(x - 5)^2(x + 3)]*[(x + 3)(2x - 3) - 2(x + 2)(2x - 2)] = 0
=> (x - 5)^2(x + 3) = 0
=> x = 5, x = -3
or
=> (x + 3)(2x - 3) - 2(x + 2)(2x - 2) = 0
=> 2x^2 + 3x - 9 - (4x^2 + 4x - 8) = 0
=> 2x^2 + 3x - 9 - 4x^2 - 4x + 8 = 0
=> -2x^2 - x - 1 = 0
=> 2x^2 + x + 1 = 0
x = [-1 +/- sqrt(1 - 8)]/4
no real solution----> there are no inflection points

EDIT: it turns out that x = 5 is not an asymptote either. Since we could have factorized the bottom and cancel the (x - 5) on top. so the domain includes 5 and 5 is not a critical point. In retrospect, life would have been a LOT easier if we factorized and simplified before doing all these operations. now that youknow the steps, factorize f(x) cancel what needs to be canceled and redo the steps yourself.

3. Originally Posted by vc15ao4
Find the area bound betweeen the curves f(x)=x^2-4x+3 and f(x)= -2x+3
first thing we do is draw the graphs. this allows us to see which is higher then which

the area bounded between them is given by int{(higher graph) - (lower graph)}dx

second thing to do is to find where the graphs intersect, so we can use these as the limits of our integration. they intersect where there formulas give the same result:

for intersection:
x^2 - 4x + 3 = -2x + 3
=> x^2 - 2x = 0
=> x(x - 2) = 0
=> x = 0, x = 2
these are the limits of our intergral. see the attachments below for the graph and solution.

4. Originally Posted by vc15ao4
The radius r of a sphere is increasing at a rate of .7 inches per minute. Find the rate of change of the volume when the radius is 4 inches. V=4/3(pieR^3)
first of all, it's "pi," "pie" is something that you eat.

let the volume be V
let the rate at which the volume is changing be dV/dt
let the rate at which the radius is changing be dr/dt

using V = (4/3)pi*r^3
we employ implicit differentiation with respect to t
=> dV/dt = 4pi*r^2 dr/dt

given: dr/dt = 0.7, r = 4
=> dV/dt = 4pi*(4)^2 * 0.7 = 44.8pi

so the volume is increasing at a rate of 44.8pi ~= 140.74 cubic inches per minute

5. Originally Posted by vc15ao4

Find the volume of the solid formed by revolving the region bounded by f(x)=4x-x^2 and g(x)=2 around the line x=5
First we find where the graphs intersect to get the limits of integration. they intersect where their formulas give the same result, that is, where they are equal.

For intersection:
4x - x^2 = 2
=> x^2 - 4x + 2 = 0
x = [4 +/- sqrt(16 - 8)]/2
=> x = (4 + sqrt(8))/2 or (4 - sqrt(8))/2
=> x = 2 + sqrt(2) or 2 - sqrt(2)

Now we will use the method of Cylindrical Shells (or the Shell method) to find the olume of revolution

See the attachments below:

the first is the diagram, the last is the solution.

6. Originally Posted by vc15ao4
Find the volume of the solid formed by revolving the region bounded by f(x)= 4x-x^2 and g(x)=2 around the line y=2
For this, we employ the disk method. we already found the limits of integration above.

Again, the first attachment is the diagram, the last is the solution