SOLUTION: Let {{{f(x)= 1/(2x^3-9)}}} and find {{{f^-1(x)}}}

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Question 1007160: Let f%28x%29=+1%2F%282x%5E3-9%29 and find f%5E-1%28x%29
Found 2 solutions by Theo, MathLover1:
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
f^-1(x) is the inverse function of f(x).

let y = f(x).

you get y = 1/(2x^3-9)

replace y with x and x with y and you get:

x = 1/(2y^3-9)

that's your inverse equation.

what's left is to solve for y if you can.

if not, go with it the way it is.

this one can be solved for y.

start with x = 1/(2y^3-9)

multiply both sides of the equation by (2y^3-9) and divide both sides of the equation by x.

you will get:

2y^3 - 9 = 1/x

add 9 to both sides of the equation to get:

2y^3 = 1/x + 9

since 1/x + 9 is the same as (1+9x)/x, the equation becomes:

2y^3 = (1+9x)/x

divide both sides of the equation by 2 to get:

y^3 = (1+9x)/(2x)

take the third root of both sides of the equation to get:

y = ((1+9x)/(2x))^(1/3)

that's your inverse equation after you have solved for y.

the grpah of the equation plus its inverse is shown below.

the graphs also includes the line y = x which they need to be a reflection of in order to be inverse equations.

the graph also includes lines y = -x + 5 and y = -x - 5 thst shows that the points are reflections about the line y = x.

that's because (x,y) on the original equation equals (y,x) on the inverse equation.

$$$








Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=+1%2F%282x%5E3-9%29
find: f%5E-1%28x%29
recall f%28x%29=+y

y=+1%2F%282x%5E3-9%29........swap x and y

x=+1%2F%282y%5E3-9%29........solve for y

x%282y%5E3-9%29=+1

2y%5E3-9=+1%2Fx

2y%5E3=+1%2Fx%2B9

y%5E3=+%281%2Fx%2B9%29%2F2

y=+root%283%2C%281%2Fx%2B9%29%2F2%29

y=+root%283%2C1%2Fx%2B9%29%2Froot%283%2C2%29

so, your inverse is: f%5E-1%28x%29=root%283%2C1%2Fx%2B9%29%2Froot%283%2C2%29