How do you find the inverse of #y=(3x-7)/(x+9)#?

Answer 1

#f^-1(x)=(9x+7)/(3-x)#

let, #y=f(x)=(3x-7)/(x+9)# then we have #x=f^-1(y)# #rArry=(3x-7)/(x+9)rArry(x+9)=3x-7rArrxy+9y=3x-7# #rArr9y+7=x(3-y)rArrx=(9y+7)/(3-y)# since #x=f^-1(y)rArrf^-1(y)=(9y+7)/(3-y)rArrf^-1(x)=(9x+7)/(3-x)#
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Answer 2

#f(x)^-1= (-9x-7)/(3+x)#

The inverse of a function switches the imput value and the output value. One easy way to solve inverse functions is by simply switching where the #x's and y's are #. So... #f(x) = (3x-7)/(x+9) # turns into # x = (3y-7)/(y+9)# Then from here on it is basic algebra. # x = (3y-7)/(y+9)# #x*(y+9) =( 3y-7)# #xy+9x = 3y -7 # #3y+xy = -9x-7# #y(3+x)=-9x-7# #f(x)^-1= (-9x-7)/(3+x)#

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Answer 3

To find the inverse of the function (y = \frac{3x - 7}{x + 9}), first, swap the roles of (x) and (y), then solve for (y). The inverse function is (x = \frac{3y - 7}{y + 9}). Now solve this equation for (y). After simplification, the result is (y = \frac{9x + 7}{x - 3}), which represents the inverse of the original function.

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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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