I need some help with math(precal)?

Question

Alright, I made a huge mistake taking honors precal and I have a couple questions on how to do questions on my homework that is due tomorrow (I can't consult my book because , just to screw with us, my teacher said she got these problems out of a different book). I don't necessarily need the answers, just some help with how to do them. (can't ask the teacher either, because they are due before school starts)

the first one I am having trouble on...
Graph the function over the interval (-1, 3) state where the function is increasing, decreasing, and constant. Also state any relative minimums or maximums to 2 decimal places.
--I get how to graph and stuff, but does it mean only include increasing/decreasing over the interval or all points?--

second....
find f^-1(x) if f(x) = (2x+1)/(x-1) x does not equal 1
--this is finding the inverse, and I can usually get these with no problem, except I have never done one where it is over an "x" equation. I can't get passed x(y-1) = 2y +1

continuing...
third - how would you set this one up:
A farmer has 2000 yards of fence to enclose a rectangular plot that borders a highway. If the farmer does not fence the side along the highway, what is the largset area that can be enclosed?
--I can't seem to set it up right--

last....
if 2^x = 3 what does 4^-x equal?

Answer 1

only over the interval

y = (2x+1)/(x-1)
y(x-1) = 2x+1
yx-y = 2x+1
yx - 2x = 1 + y
x(y - 2) = 1 + y
x = (1+y)/(y-2)
so x = f^-1(y) = (1+y)/(y-2)
so now put x's in place of the y's
gives
f^-1(x) = (1+x)/(x-2)

let y be the side parallel to the highway and x be the two sides perpendicular to the highway, then
length = 2000 = 2x + y
so y = 2000 - 2x
area = A = xy
so
A = x(2000 - 2x) = 2000x - 2x^2
now find the maximum
x = -b/2a = -2000/2*(-2) = 500
so y = 2000 - 2*500 = 1000
gives
Amax = 500*1000 = 500000 yards^2

2^x = 3
so
4^-x = 1/4^x = 1/(2^2)^x = 1/2^(2x) = 1/(2^x)^2 = 1/3^2 = 1/9

the end
.

Answer 2

> does it mean only include increasing/decreasing over the interval or all points?

I admint it's vague, but I take it to mean: List each interval where it's increasing & each interval where it's decreasing; whether or not those intervals are within the (-1,3) interval that you graphed. Look at it this way: If you include the full domain of the function, then you're sure to include the (-1,3) interval too, so you can't go wrong.

> I have never done one where it is over an "x" equation.

Don't worry about what letter is used; that doesn't make any difference, really. However, it's sometimes easier to solve if you make some substitution, like using the variable "y" in place of "f(x)"

f(x) = (2x+1)/(x-1)
y = (2x+1)/(x-1)

Once you've got it at that point, here are the steps for finding the inverse:

1) Make a new equation in which the x's and y's are swapped:
x = (2y+1)/(y-1)
(I think you got that far already)

2) Solve for "y" (or whatever was the new variable you introduced).

I think this is the part you're having trouble with. But it's fairly basic algebra:

x = (2y+1)/(y-1)

Multiply both sides by (y-1):
x(y-1) = 2y+1

Distributive law on left side:
xy - x = 2y + 1

Add "x" to both sides:
xy = 2y + 1+ x

Subract "2y" from both sides:
xy - 2y = 1 + x

Distributive law:
(x - 2)y = 1 + x

Divide both sides by (x - 2):
y = (1 + x) / (x - 2)

> third - how would you set this one up:

1) The thing you're trying to maximize is "area", so write a function that looks like "area = (something)". To figure out what the "(something)" is, you need to think about some single variable that you have control over, which will affect the total area. For example, choose "w", the width of the part of the fence that runs parallel to the road. Given that the total fence length is 2000, figure out a formula with "w" in it, that would give you the total area.

2) Once you have "area=(something with 'w' in it)", find the value of 'w' that gives you the maximum 'area'.

Answer 3

For your first question, just simply graph the increasing/decreasing/constant function within that intervals, not at all point.

For the 2nd question, you want to first take the inverse, then solve for x. hey, you can consult some of the pre-cal information online at well. Just google this, and I am sure you will find more information.

good luck!

Answer 4

first...you probably need over all points and the end behavior
seond...y=(2x+1)/(x-1) switch x and y...
x=(2y+1)/(y-1) mult by y-1
x(y-1)=2y+1 distribute
xy-x=2y+1 set it all equal to 1
-2y-xy-x=1 factor out y and add x
y(x-2)=x+1 divide by x-2
y=(x+1)/(x-2)
last...2^x = 3 log base 2 of each side
x=logbase2 (3)

Answer 5

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