One to one function h is defined by h(x)=(6x)/(-8+7x). Find h^-1, the inverse of h. Then, give the domain and range of h^-1 using interval notation.

Question

johnweldon1993 · Answer

Do you know how to find the inverse?

johnweldon1993 · Answer

You begin by switching the 'x's and the 'y's

So your equation

$$y = \frac{ 6x }{ -8 + 7x }$$

Will become

$$x = \frac{ 6y }{ -8 + 7y }$$

Now you just solve for 'y' again...

johnweldon1993 · Answer

So multiply both sides by -8 + 7y to get rid of that fraction on the right hand side

$$x(-8 + 7y) = 6y$$

Distribute out that 'x'

$$-8x + 7xy = 6y$$

Now you want to solve for 'y' so move everything with a 'y' to 1 side

$$6y - 7xy = -8x$$

Factor out a 'y' from the left hand side

$$y(6 - 7x) = -8x$$

And finally to solve for 'y' divide both sides of the equation by 6 - 7x

$$y = \frac{ -8x }{ 6 - 7x }$$

That is your inverse....can you tell me the domain or range?

anonymous · Answer

i dont understand how to find the domain and range

johnweldon1993 · Answer

To find the domain...you find out what values of 'x' will make your denominator = 0 so

$$-7x + 6 = 0$$

Can you solve that equation for 'x'...?

anonymous · Answer

x would be 6/7 how do you find the range then?

johnweldon1993 · Answer

Well the range would be all real numbers not equal to that number we just found...the domain...

anonymous · Answer

how would i write that? -infinity, infinity?