Question

Let f(x) = ln(9 − x).

(a) Find the domain of f.
−infinity < x < infinity
x ≤ 9
x ≥ 9
x < 9
x > 0

(b) Find the range of f.
y > 0
y > 4
−infinity < y < infinity
y ≥ 4
y < 9


(b) Find f^−1(x).
f^−1(x) =

Answer 1

for a it should be x>0

Answer 2

wait i mean x<=9

Answer 3

ln(0) is also undefined

Answer 4

huh what happened to his answer??

Answer 5

ln(x) domain is all x > 0
so ln(9-x) domain is when 9-x> 0 -x> -9
x<9

Answer 6

y = ln(9-x) so
e^y = 9-x
so e^y-9 = -x
x = 9-e^y
so the range is all real numbers

Answer 7

and the inverse I already showed but rename x to y and y to x
y = 9-e^x

Answer 8

can u explain this to me, my teacher is horrible :(


y = ln(9-x) so
e^y = 9-x
so e^y-9 = -x
x = 9-e^y

Answer 9

The way I find the range of something is first take the inverse, then I see what are the domain restrictions on the inverse. this tells me the range.

Answer 10

so like y = 1/x has a domain of all x such that x != 0 and the range...
y = 1/x
so x = 1/y now the range is all the values of y, and we see clearly that y cant be 0, so the range is all numbers x such that x != 0

Answer 11

where do u get the e^y tho??

Answer 12

y = ln(9-x)
e^y = e^ln(9-x)
e^y = 9-x
make sense?

Answer 13

if y = lnx then x = e^y by definition of ln(x)