Question

f(x)=x-3, g(x)=2x-8 and h(x)=x^(2)-2
a) what is f(x)·g(x)?
is it (x-3)·(2x-8)
b) f(x)/g(x) what is the domain of the function h(x)=f(x)/g(x)

Answer 1

you're right for f(x).g(x) for f(x)/g(x) use (x-3)/(2x-8) and see where it loses the definition of function

Answer 2

just see (2x-8)=0 find x ...domain will be all real number except x= 4

Answer 3

i get the (2x-8)=0 and x=4 part, but how do i know where it loses the definition of a function
actually, what is the definition of a function

Answer 4

Basically, when x = 4, 2x-8 is zero, which causes a division by zero error

Answer 5

Which is why you have to exclude this value from the domain.

Answer 6

well if you put x=4 in the function (x-3)/(2x-8) you'll have infinity that means your function is not defined at x=4 ...so you've to exclude x=4

Answer 7

at rest of the points your function exists

Answer 8

so then, how do i find the domain of that function?

Answer 9

all real number greater than 4?

Answer 10

no, all real numbers but x = 4

Answer 11

oh, thank you :)

Answer 12

or I should say \(\Large x \neq 4\)

Answer 13

np