ACT math question!!?

Question

This is probably only an algebra question but I can't understand it

"For all x>3, (3x - x^2) / (x^2 + 3x -18) = ?"

Thanks, and also had one more question...
"If the function f satisfies the equation f(x+y) = f(x) + f(y) for every pair of real numbers x and y, what is(are) the possible value(s) of f(0)?"
Options:
any real number
any positive real number
0 and 1 only
1 only
0 only





Thanks for any and all help!! =)

Answer 1

For all x>3, (3x - x^2) / (x^2 + 3x -18) =
x(3 - x) / (x + 6)(x - 3) then factor a -1 out of the (3 - x) term and cancel it with the bottom one.
= -x / (x + 6), x>3

For the second question,
(0,0) is a possible pair which would give f(0) = f(0) + f(0) =
2f(0)....this is only true for f(0) = 0, so 0 only.

Answer 2

f(0)=f(0+0)=f(0)+f(0);
cancel out f(0) on each side,
0=f(0);

For x>3, we get
x^2>9, 3x>9
x^2+3x-18>9+9-18=0, denominator is positive
x>0, 3-x<0
3x-x^2=x(3-x)<0, numerator is negative
the fraction must be negative

Further, as x is near 3(slightly larger than it),
the numerator is near zero,
the value of the fraction approaches infinity.
As x increases without bound, the fraction approaches -1.

Since the derivative of the function is always positive at x>3, the fraction cannot have a local max or min, it is strictly increasing.

Answer 3

Factor both sides of the fraction, and you get

x(3-x)/[(x+6)(x-3)]

3-x and x-3 are opposites, so they cancel if you multiply one of them by -1. Then you have:

-x/(x+6) for your solution.

_____________________________

On the second question, I think the answer must be 0 only.

f(x+0) = f(x) + f(0)

But x + 0 = x, so this means that

f(x) = f(x) + f(0), so f(0) has to = 0 for that to be true.

Answer 4

The x>3 stipulation is put in since at x=3, the denominator "blows up". You are also free to stipulate any domain restriction you care to establish, so there really isn't any pressing reason why the expression couldn't exist for x<3.

Answer 5

if you factor out a -x from the top part of the division and factor the bottom of it, you end up with -x(x-3)/(x+6)(x-3). The (x-3)'s cancel, leaving you with -x/x+6.

Answer 6

well i got the first one

3x - x^2
--------------
x^2 + 3x - 18


distribute -x out of the top

and factor the bottom

-x (-3 + x)
--------------
(x-3) (x+6)

then cancel out

-x
------
x+6