How do you find the inverse of #h(X)= 5 / (2x + 3)# and is it a function?

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Answer 1 · 2016-03-20T03:49:50

The inverse of a function can be found algebraically by switching the x and y values

#y = 5/(2x + 3)#

#x = 5/(2y + 3)#

#x(2y + 3) = 5#

#2y + 3 = 5/x#

#2y = (5 - 3x)/x#

#y = (5 - 3x)/(2x)#

#h^-1(x) = (5 - 3x)/(2x)#

Here are a few things to remember when finding the inverse of a function:

The y must be isolated (all alone on one side of the equation).
Don't forget the #h^-1(x) notation. I have been docked marks before from forgetting to include this element in my answer.
The inverse of a function can be found graphically by reflecting the original function over the line #y = x#

The first graph below is of the original function. The second is of the inverse.

graph{y = 5/(2x + 3) [-10, 10, -5, 5]}

graph{y = (5 - 3x)/(2x) [-4.933, 4.933, -2.466, 2.467]}

Practice exercises:

Indicate the inverses of the following functions, and then state whether or not they are functions.

a) #ƒ(x) = 2x + 7#

b) #h(x) = 3x^2 - 5x + 1#

c) #j(x) = sqrt(2x - 4) + 5#

d) #k(x) = (3x - 2)/(5x + 4)#

#2.# The following graph is of function #ƒ^(-1)(x)#. State the equation of #ƒ(x)#

graph{y = 1/3x + 2 [-9.51, 9.51, -4.755, 4.755]}

Good luck!