find the inverse of the function (x^3-8)/x-2?

Question

topic;Mathematics
Sub topic;Functions

Answer 1

(x-2)^(1/2) - 1


Inverse of function is not 1 / F(x)
Its the reflection of the function across the line y = x

Answer 2

to discover an inverse function merely exchange the x and y and rearrange for y. (x+2)^2=(x+2)(x+2) =x^2 + 4x + 4 to bypass a fragment to the different side merely cases by utilising the inverse i.e x-4=3/8x 8/3(x-4)=x or 8(x-4)=3x relies upon the way you desire

Answer 3

f(x) = (x^3 - 8)/(x - 2)

First, obtain the domain and range.
Domain: (-infinity, 2) U (2, infinity)
Range: [3, 12) U (12, infinity)

We do this because the domain and range will be swapped upon finding the inverse.

Make f(x) = y.

y = (x^3 - 8)/(x - 2)

Swap y and x, and solve for x.

x = (y^3 - 8) / (y - 2)
x(y - 2) = (y^3 - 8)
x(y - 2) = (y - 2)(y^2 + 2y + 4)
x = y^2 + 2y + 4
x = y^2 + 2y + 1 + 3
x = (y + 1)^2 + 3
x - 3 = (y + 1)^2

Therefore,

(y + 1)^2 = x - 3
y + 1 = +/- sqrt(x - 3)
y = -1 +/- sqrt(x - 3)

But, since the domain is [3, 12) U (12, infinity) and the range is (-infinity, 2) U (2, infinity), we take only the positive result.

y = -1 + sqrt(x - 3)

Therefore,

f^(-1)(x) = -1 + sqrt(x - 3),
for x in [3, 12) U (12, infinity).

Answer 4

y = (x^3 - 8) / (x - 2)
y = (x^3 - 6x^2 + 12x - 8 + 6x^2 - 12x) / (x - 2)
y = ((x - 2)^3 + 6x(x - 2)) / (x - 2)
y = (x - 2)^2 + 6x
y - 4 = x^2 + 2x
y - 4 + 1 = x^2 + 2x + 1
(y - 3) = (x + 1)^2
x + 1 = (y - 3)^(1/2)
x = ± (y - 3)^(1/2) - 1
swapping variables,
y = ± (x - 3)^(1/2) - 1

Answer 5

x^3-8/x-2 = (x-2)(x^2+2x+4)/x-2 = x^2+2x+4

Reverse = 1/x^2+2x+4

Answer 6

first x^3 - 8 = (x-2) (x^2 + 2x + 4)

F^(-1) =
(x-2)/{(x - 2)(x^2 + 2x +4)}
= 1/(x^2 + 2x +4)

Answer 7

inverse is NOT one over.. that is reciprocal

inverse means switch x<====>y and solve!

I SWEAR!