# Inverse Of X^3

How Do You Find The inverse function of g(x)=x^3,f(x)=1/8x-3?

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For a function y=f(x), the inverse function g(x) is such that g(f(x))=1.

For the functions you have provided:

- f(x)=x^3

y=x^3

Take the cube root of y and x^3, => y^(1/3)=x

Therefore the inverse function is x^(1/3).

**f(x)= x^(1/3) is the inverse function of f(x)=x^3**

- f(x)=1/(8x-3)

y=1/(8x-3)

=>(8x-3)=1/y

=>8x=(1/y)+3

=> x=[(1/y)+3]/8=(3y+1)/8y

**The inverse function of f(x)=f(x)=1/(8x-3) is f(x)=(3x+1)/8x**

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We'll find the inverse of the function f(x).

f(x)=1/(8x-3)

We'll note f(x) as y, so, y = f(x) = 1/(8x-3)

Now, we'll multiply the denominator (8x-3) by y:

y* (8x-3) = 1

We'll remove the brackets:

8xy - 3y = 1

We'll isolate y to the left side. For this reason, we'll add 3y both sides:

8xy = 1 + 3y

We'll divide by 8y both sides:

x = (1+3y)/8y

The expression of the inverted function is:

f(y) = (1+3y)/8y

Conventionally, a function has as variable x, so, we'll re-write the inverse function:

**f^-1(x) = (1+3x)/8x**

To determine the inverse function of g(x), we'll cover the same ground.

First, we'll note the function g(x) = y.

g(x) = y = x^3

We'll keep y = x^3.

We'll raise to the power (1/3) both sides:

y^(1/3) = x

The inverse function of g(x) is:

**g^-1(x) = x^(1/3)**

To find the inverse of g(x) = x^3 and f(x) = 1/(8x-3)

Solution:

Let y = x^3.

We take cube roots.

y^(1/3) = x.

So x=y^(1/3).

Swap xand y

y = x^(1/3) is the inverse of g(x) = x^3.

(2) f(x) = 1/(8x-3).

Let y= 1/(8x-3)

Make x subject. Or try solve for x, by multiplying by (x-3) and dividing by y:

8x-3 = 1/y.

8x = 1/y +3

8x = (3y+1)/y.

x = (3y+1)/8y

Swap x and y:

y = (3x+1)/8x is the inverse of f(x) = y = 1/(8x-3)