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

Answer 1

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

#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#
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Answer 2

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

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#
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Answer 3

The inverse function of ( h(x) = 3 - \frac{x + 4}{x - 7} ) is denoted as ( h^{-1}(x) ). To find the inverse function, we swap the roles of ( x ) and ( y ) and solve for ( y ).

After finding the inverse function, to evaluate ( h^{-1}(9) ), we substitute ( x = 9 ) into the inverse function and solve for ( y ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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