How Do You Find The inverse function of g(x)=x^3,f(x)=1/8x-3?
For a function y=f(x), the inverse function g(x) is such that g(f(x))=1.
For the functions you have provided:
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
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).
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)
Let y = x^3.
We take cube roots.
y^(1/3) = x.
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)