What is the inverse function of #h(x)= 3-(x+4)/(x-7)# and how do you evaluate #h^-1(9)#?

2 Answers
Jun 11, 2015

Answer:

#h^(-1)(y)=(7y-25)/(y-2)# where #y!=2#.
#h^(-1)(9)=38/7#

Explanation:

#h(x)= 3-(x+4)/(x-7)# having #x!=7#
I find the common denominator and sum all together:
#h(x)=(3x-21)/(x-7)-(x+4)/(x-7)#
#h(x)=(2x-25)/(x-7)#
Now I simplify making the division of the 2 polynomials and I obtain:
#"quotient"=2, "reminder"=-11#
So I can write the function as:
#h(x)=2(x-7)/(x-7)-11/(x-7)#
#h(x)=2-11/(x-7)#.

Now, the question is how to find the inverse function? Firstly, I try to isolate x:
#(x-7)(h(x)-2)=-11#
#(x-7)=-11/(h(x)-2)#
#x=-11/(h(x)-2)+7#
Therefore we rewrite better the function:
#h^(-1)(y)=-11/(y-2)+7=(7y-14-11)/(y-2)=(7y-25)/(y-2)#.
So we can state that:
#h^(-1)(y)=(7y-25)/(y-2)# where #y!=2#.

If we want to find #h^(-1)(9)#:
#h^(-1)(9)=(7*9-25)/(9-2)=(63-25)/7=38/7#

Jun 11, 2015

Answer:

The inverse function is #h^(-1)(x) = (7x-25)/(x-2)#.
#h^(-1)(x) = 38/7#

Explanation:

Since the original function is not that complex, you can determine its inverse function faster by solving the function for #x# and switching the result for #h(x)#.

#h(x) = 3 - (x+4)/(x-7)#

#h(x) = (3(x-7)- (x+4))/(x-7) = (3x - 21 -x - 4)/(x-7)#

#h(x) = (2x - 25)/(x-7)#

Solve this form of the equation for #x# to get

#h^(x) * (x-7) = 2x-25#

#x * h(x) - 7 * h(x) * 7 = 2x-25#

#x * h(x) - 2x = 7 * h(x) - 25#

#x( h(x) - 2) = 7 * h(x) - 25#

#x = (7 * h(x) - 25)/(h(x) - 2)#

Once you isolate #x#, simply switch #h(x)# for #x# to get the inverse function

#h^(-1)(x) = (7x - 25)/(x-2)#

To evaluate #h^(-1)(9)#, simply substitute #x# with #9# to get

#h^(-1)(9) = (7 * 9 - 25)/(9-2) = 38/7#