domain and range of 5x-4/(3+2x).. work shown need help!

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johnq2k7

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Given the function f(x)= (3x+4)/(5-2x) find the domain and range of the inverse function of f(x)

work shown:

it was proven to be one to one since f(x1)= f(x2)

found inverse f(x) to be (5x-4)/(3+2x), i believe this is correct

therefore the domain of the inverse function is the range of f(x)

therefore the range of the inverse function is the domain of f(x)

therefore the range of the inverse function is (-inf, 5/2), (5/2, inf)

since the domain of f(x) can't be 5/2 since 5-2x cannot equal 0 therefore x=0

i believe i got the domain of the inverse function correctly

however what is the range of the inverse function, without graphically determining it
 
johnq2k7 said:
Given the function f(x)= (3x+4)/(5-2x) find the domain and range of the inverse function of f(x)

work shown:

it was proven to be one to one since f(x1)= f(x2)

found inverse f(x) to be (5x-4)/(3+2x), i believe this is correct

therefore the domain of the inverse function is the range of f(x)

therefore the range of the inverse function is the domain of f(x)

therefore the range of the inverse function is (-inf, 5/2), (5/2, inf)

since the domain of f(x) can't be 5/2 since 5-2x cannot equal 0 therefore x=0 ? That statement does not make sense

i believe i got the domain of the inverse function correctly

however what is the range of the inverse function, without graphically determining it - you answered it above
Click to expand...

The original function has a vertical asymptote at x = 5/2

Inverse function has a horizontal asymptote ? y = 5/2
 
Subhotosh Khan said:
johnq2k7 said:
Given the function f(x)= (3x+4)/(5-2x) find the domain and range of the inverse function of f(x)

work shown:

it was proven to be one to one since f(x1)= f(x2)

found inverse f(x) to be (5x-4)/(3+2x), i believe this is correct

therefore the domain of the inverse function is the range of f(x)

therefore the range of the inverse function is the domain of f(x)

therefore the range of the inverse function is (-inf, 5/2), (5/2, inf)

since the domain of f(x) can't be 5/2 since 5-2x cannot equal 0 therefore x=0 ? That statement does not make sense

i believe i got the domain of the inverse function correctly

however what is the range of the inverse function, without graphically determining it - you answered it above
Click to expand...

The original function has a vertical asymptote at x = 5/2

Inverse function has a horizontal asymptote ? y = 5/2
Click to expand...


sorry i worded answers incorrectly

i meant since the function f(x) cannot be x=5/2 therefore
the range of the inverse function is the domain of f(x)
therefore the range of the function is (-inf, 5/2), (5/2,inf)

however.. i needed help wit the domain of the inverse function
since the domain of the inverse function is the range of the function f(x)

i found a horizontal asymptote of y=-1 for the function f(x) ... this may be incorrect

therefore the domain of the inverse function is (-inf,-1), (-1, inf)

is this correct?... sorry for the incorrect wording used before
 
[quote="johnq2k7]

sorry i worded answers incorrectly

i meant since the function f(x) cannot be x=5/2 therefore
the range of the inverse function is the domain of f(x)
therefore the range of the function is (-inf, 5/2), (5/2,inf)

however.. i needed help wit the domain of the inverse function
since the domain of the inverse function is the range of the function f(x)

i found a horizontal asymptote of y=-1 for the function f(x) How - please show work


... this may be incorrect

therefore the domain of the inverse function is (-inf,-1), (-1, inf)

is this correct?... sorry for the incorrect wording used before[/quote]
 
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