Domain and range of the function help needed work shown

[johnq2k7]

domain and range of the function help needed.. work shown!

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 may have made some incorrect assumptions for the range of the inverse function correctly

however what is the range of the inverse function, without graphically determining it
and is my domain for the inverse function correct

please help me with this problem

 

[quantumdude]

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)

The function is 1-1, but the reason you cite makes no sense at all. A function $f:D\rightarrow R$ is 1-1 if and only if, for all $x_1,x_2\in D$, $x_1\neq x_2$ implies $f(x_1)\neq f(x_2)$.

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

There's an easy way to check. If $f^{-1}(f(x))=x$ then you've found the correct inverse.

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

Correct.

however what is the range of the inverse function, without graphically determining it
and is my domain for the inverse function correct

I'm confused. You gave us the (correct) range of the inverse function, and you didn't give the domain.

 

[johnq2k7]

The function is 1-1, but the reason you cite makes no sense at all. A function $f:D\rightarrow R$ is 1-1 if and only if, for all $x_1,x_2\in D$, $x_1\neq x_2$ implies $f(x_1)\neq f(x_2)$.



There's an easy way to check. If $f^{-1}(f(x))=x$ then you've found the correct inverse.



Correct.



I'm confused. You gave us the (correct) range of the inverse function, and you didn't give the domain.

I made a mistake i meant f(x1) is not equal to f(x2) therefore it is one to one...

and i found the range of the inverse which is the domain of the function to be (-inf, 5/2), (5/2,inf)

however, although the domain of the inverse is equal to the range of the function.. i found the domain of the inverse function to be (-inf,-1), (-1,inf)... but i believe this is wrong...


so you have concluded that my inverse functions's range is correct.. but i need help with the domain of the inverse function.. please help!

 

[quantumdude]

I've just checked your inverse function, and it's right. To find its domain just do exactly what you did when you found the original function's domain: determine the value of x that makes the denominator zero. That's the one value of x that's excluded from the domain.

 

[johnq2k7]

I've just checked your inverse function, and it's right. To find its domain just do exactly what you did when you found the original function's domain: determine the value of x that makes the denominator zero. That's the one value of x that's excluded from the domain.

ok.. then i get 3+2x=0

therefore x= -3/2

therefore is the domain of the inverse function, (-inf, -3/2), (-3/2, inf)

i believe this should be correct.. thanks for the help

 

[quantumdude]

Yes, that's right.

 

[Mark44]

Tom,
What johnq2k7 said in his first post does make sense, although he didn't complete his thought.

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

f(x1) = f(x2) ==> x1 = x2 is an equivalent way (the contrapositive) to say
x1 $\neq$ x2 ==> f(x1) $\neq$ f(x2)