Question

f(x) = x2 - 16 and g(x) = x+4. Find f/g and its domain.

Answer 1

A.
x - 4; all real numbers except x \[\neq-4\]

B.
x + 4; all real numbers except x \[\neq-4\]

C.
x + 4; all real numbers except x \[\neq4\]

D.
x - 4; all real numbers except x\[\neq4\]

Answer 2

I am not good at math :\

Answer 3

I'm soooo sorry but I'm not good at math either. :s
@suckerofmath can you help? c:

Answer 4

I have no idea sorry.

Answer 5

She is intelligent in an inarticulate manner ^.^
Just kidding Tea :p

Answer 6

Lol ohh wow

Answer 7

So first we know that a^2 - b^2 = (a -b) * (a + b)

==> x^2 - 16 = x^2 - 4^2 = (x+4) * (x - 4)

Since we are finding domain of f/g = (x+4)*(x - 4)/(x+4) = (x - 4)

So Domain of (x - 4) is ??? this function f/g is exist for all real values

therefore Domain is R

Answer 8

I have no idea here

Answer 9

Okay We are finding domain of f/g = x^2 - 16 /x+4 right ?

Answer 10

yes

Answer 11

& we know that a^2 - b^2 = (a -b) * (a + b)

So we can write x^2-16 as =x^2 - 4^2 = (x- 4) * (x + 4) okay ?

Answer 12

ok

Answer 13

So we have got f/g = (x + 4) * (x - 4) / (x +4) right ?

Answer 14

yup

Answer 15

Now we cancel common factor fro numerator & denominator

So we will get f/g = (x + 4)*(x-4)/(x+4) = (x - 4) Ok ?

Answer 16

ok

Answer 17

Answer is D ?

Answer 18

Now we have f/g = (x - 4) & we are finding domain
Domain is the set of values for which function is defined or function make sence

So Here at any value of x our function f/g is defined , that is if we put any real number instead of x we get some answer that is function is defined at all real values

Hence Domain = R

Answer 19

Is D the answer ?

Answer 20

no please check it the function is undefined when denominator is zero

our original function f / g = x^2 - 16 / (x + 4) so at what value denominator become zero ?

Answer 21

A ?

Answer 22

thanks

Answer 23

u r welcome :)

Answer 24

#Shay17 #tgawade was the answer to this problem A? I'm so lost right now