inverse function for f(x) = x^3 +12x^2 + 48x + 61
- Thread starter mikeal27
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Re: inverse functions
If you are given f(x) = x[sup:35t33gui]3[/sup:35t33gui] + 12x[sup:35t33gui]2[/sup:35t33gui] + 48x + 61, then let y = f[sup:35t33gui]-1[/sup:35t33gui](x). Then, you would just replace all the x's with y's, and the f(x) with x's. You might not be able to solve for f[sup:35t33gui]-1[/sup:35t33gui](x) but at least you got some algebraic representation for it.
As for the approximate graphical representation, think of all the (x,y) coordinates that satisfy f(x). What would the coordinates be for f[sup:35t33gui]-1[/sup:35t33gui](x) in terms of x and y? If you know this, you can pick out a few points and attain a rough sketch of the inverse of f(x). Also, keep in mind that f[sup:35t33gui]-1[/sup:35t33gui](x) is the reflection of f(x) over the line y = x.
If you are given f(x) = x[sup:35t33gui]3[/sup:35t33gui] + 12x[sup:35t33gui]2[/sup:35t33gui] + 48x + 61, then let y = f[sup:35t33gui]-1[/sup:35t33gui](x). Then, you would just replace all the x's with y's, and the f(x) with x's. You might not be able to solve for f[sup:35t33gui]-1[/sup:35t33gui](x) but at least you got some algebraic representation for it.
As for the approximate graphical representation, think of all the (x,y) coordinates that satisfy f(x). What would the coordinates be for f[sup:35t33gui]-1[/sup:35t33gui](x) in terms of x and y? If you know this, you can pick out a few points and attain a rough sketch of the inverse of f(x). Also, keep in mind that f[sup:35t33gui]-1[/sup:35t33gui](x) is the reflection of f(x) over the line y = x.
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Re: inverse functions
To find the graph of the inverse (which is a function, by the way), plot a bunch of points for f(x), reverse the x- and y-values, and then plot the new (reversed) points. These new points should allow you to sketch f[sup:8w3uiy49]-1[/sup:8w3uiy49](x).
To find the "formula", you'll have to be clever. If you have a quadratic function and need to find the inverse, you know it's easy if it can be put into the form "y = a(x - h)[sup:8w3uiy49]2[/sup:8w3uiy49] + k" or just "y = a(x - h)[sup:8w3uiy49]2[/sup:8w3uiy49]":
. . . . .f(x) = y = 3(x - 2)[sup:8w3uiy49]2[/sup:8w3uiy49], x < 2
Switch the variables:
. . . . .x = 3(y - 2)[sup:8w3uiy49]2[/sup:8w3uiy49]
Solve for "y=":
. . . . .x / 3 = (y - 2)[sup:8w3uiy49]2[/sup:8w3uiy49]
. . . . .+/-sqrt[x / 3] = y - 2
. . . . .+/-sqrt[x / 3] + 2 = y
Since x < 2, the original function had been the left-hand side of the parabola, so the inverse is the bottom half of the square root:
. . . . .f[sup:8w3uiy49]-1[/sup:8w3uiy49] = -sqrt[x / 3], x > 0
If the quadratic is "messy", you have to "complete the square" before you can find the inverse (or else get messy with the Quadratic Formula, and -- trust me on this -- you do not want to mess with any "cubic" formulas!!). Can we "complete the cube" here?
Returning to your exercise: You know that a cubed binomial is of the form:
. . . . .(x + b)[sup:8w3uiy49]3[/sup:8w3uiy49] = x[sup:8w3uiy49]3[/sup:8w3uiy49] + 3x[sup:8w3uiy49]2[/sup:8w3uiy49]b + 3xb[sup:8w3uiy49]2[/sup:8w3uiy49] + b[sup:8w3uiy49]3[/sup:8w3uiy49]
Can we rearrange your cubic to "fit" this?
. . . . .x[sup:8w3uiy49]3[/sup:8w3uiy49] + 12x[sup:8w3uiy49]2[/sup:8w3uiy49] + 48x + 61
. . . . .x[sup:8w3uiy49]3[/sup:8w3uiy49] + 3(x[sup:8w3uiy49]2[/sup:8w3uiy49])(4) + 3(x)(16) + 61
. . . . .x[sup:8w3uiy49]3[/sup:8w3uiy49] + 3(x[sup:8w3uiy49]2[/sup:8w3uiy49])(4) + 3(x)(4[sup:8w3uiy49]2[/sup:8w3uiy49]) + 61
What then should "b" equal? What is the cubed-binomial form? Can you then solve for the cube-root inverse function? :wink:
Eliz.
To find the graph of the inverse (which is a function, by the way), plot a bunch of points for f(x), reverse the x- and y-values, and then plot the new (reversed) points. These new points should allow you to sketch f[sup:8w3uiy49]-1[/sup:8w3uiy49](x).
To find the "formula", you'll have to be clever. If you have a quadratic function and need to find the inverse, you know it's easy if it can be put into the form "y = a(x - h)[sup:8w3uiy49]2[/sup:8w3uiy49] + k" or just "y = a(x - h)[sup:8w3uiy49]2[/sup:8w3uiy49]":
. . . . .f(x) = y = 3(x - 2)[sup:8w3uiy49]2[/sup:8w3uiy49], x < 2
Switch the variables:
. . . . .x = 3(y - 2)[sup:8w3uiy49]2[/sup:8w3uiy49]
Solve for "y=":
. . . . .x / 3 = (y - 2)[sup:8w3uiy49]2[/sup:8w3uiy49]
. . . . .+/-sqrt[x / 3] = y - 2
. . . . .+/-sqrt[x / 3] + 2 = y
Since x < 2, the original function had been the left-hand side of the parabola, so the inverse is the bottom half of the square root:
. . . . .f[sup:8w3uiy49]-1[/sup:8w3uiy49] = -sqrt[x / 3], x > 0
If the quadratic is "messy", you have to "complete the square" before you can find the inverse (or else get messy with the Quadratic Formula, and -- trust me on this -- you do not want to mess with any "cubic" formulas!!). Can we "complete the cube" here?
Returning to your exercise: You know that a cubed binomial is of the form:
. . . . .(x + b)[sup:8w3uiy49]3[/sup:8w3uiy49] = x[sup:8w3uiy49]3[/sup:8w3uiy49] + 3x[sup:8w3uiy49]2[/sup:8w3uiy49]b + 3xb[sup:8w3uiy49]2[/sup:8w3uiy49] + b[sup:8w3uiy49]3[/sup:8w3uiy49]
Can we rearrange your cubic to "fit" this?
. . . . .x[sup:8w3uiy49]3[/sup:8w3uiy49] + 12x[sup:8w3uiy49]2[/sup:8w3uiy49] + 48x + 61
. . . . .x[sup:8w3uiy49]3[/sup:8w3uiy49] + 3(x[sup:8w3uiy49]2[/sup:8w3uiy49])(4) + 3(x)(16) + 61
. . . . .x[sup:8w3uiy49]3[/sup:8w3uiy49] + 3(x[sup:8w3uiy49]2[/sup:8w3uiy49])(4) + 3(x)(4[sup:8w3uiy49]2[/sup:8w3uiy49]) + 61
What then should "b" equal? What is the cubed-binomial form? Can you then solve for the cube-root inverse function? :wink:
Eliz.
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mikeal27 said: