Help me with function?

Question

a)F(x)=1/(x-2)
b)F(x)=square root(3-x) - 4
Determine domain and range.show your steps

Answer 1

a) D: The denominator x-2 can't be zero because if it could, the function would be undefined. So if you transpose -2 to the other side, the answer would be x=2. So the domain is "all real numbers except 2"
R: F(x) is y. Using the equation y=1/(x-2), you change solve to find x.
y=1/(x-2)
Using cross multiplication, xy-2y=1
Transposing, xy=1+2y
Dividing both sides by y, x=(1+2y)/y
The denominator y can't be zero. So the range is "all real numbers excepet 0"

b? Steps are long. Sorry, can't answer.

Answer 2

a) domain is all real except x = 2 (denominator cannot be 0), range is all real

b) domain is 3-x > 0 or x < 3, and range is -4

Answer 3

sqrt. (3-x)-4=1/(x-2)
1/[sqrt. (3-x)-4]=x-2
1/(3-x+16)=(x-2)^2
1/(3-x+16)=x^2-4x+4
1/(19-x)=x^2-4x+4
(19-x)(x^2-4x+4)-1=0
19x^2-76x+76-x^3+4x^2-4x-1=0
by combining like terms u get:
75-80x+23x^2-x^3=0
-x^3+23x^2-80x+75=0
hopefully there will vbe an equation for solving for x, but iv never heard of 1 that goes until x^3. u shud try math.com, while i research the matter.
bye 4 now. i will edit my answer if i have found the formula.

Answer 4

This is pretty basic Calculus. If you are in High School, ask your teacher for help. If you are in college, good luck in life.