document.write( "Question 258498: What is the inverse of the function f(x)=64x^3?
\n" ); document.write( "f^-1(x)=_________ \n" ); document.write( "

Algebra.Com's Answer #190322 by jsmallt9(3758) $\"\"$ $\"About$

You can put this solution on YOUR website!
To find the inverse of a function:

Repalce the function notation with \"y\"
Swap the x's and y's in your equation. This swapping of the x and y changes to equation from the equation of the function to the equation of the inverse.
Solve the inverse equation for y, if possible.
If successful in solving the inverse for y, then the inverse is a function, too. Replace the y with the notation for inverse functions.

\n" ); document.write( "Let's see how this works:
\n" ); document.write( " $\"f%28x%29+=+64x%5E3\"$
\n" ); document.write( "1. Replace the function notation with \"y\":
\n" ); document.write( " $\"y+=+64x%5E3\"$
\n" ); document.write( "2. Swap the x's and y's:
\n" ); document.write( " $\"x+=+64y%5E3\"$
\n" ); document.write( "This is the equation for the inverse relation.
\n" ); document.write( "3. Solve the inverse for y, if possible. We'll start by dividing both sides by 64:
\n" ); document.write( " $\"x%2F64+=+y%5E3\"$
\n" ); document.write( "Next we'll find the cube root of each side:
\n" ); document.write( " $\"root%283%2C+x%2F64%29+=+root%283%2C+y%5E3%29\"$
\n" ); document.write( "Using the property of radicals, $\"root%28a%2C+p%2Fq%29+=+root%28a%2C+p%29%2Froot%28a%2C+q%29\"$ , on the left side and simplifying the cube root on the right side we get:
\n" ); document.write( " $\"root%283%2C+x%29%2Froot%283%2C+64%29+=+y\"$
\n" ); document.write( "Since $\"root%283%2C+64%29+=+4\"$ , this simplifies to:
\n" ); document.write( " $\"root%283%2C+x%29%2F4+=+y\"$
\n" ); document.write( "4. If successful in solving for y, then replace the y with the notation for inverse functions:
\n" ); document.write( " $\"root%283%2C+x%29%2F4+=+f%5E%28-1%29%28x%29\"$ or $\"%281%2F4%29root%283%2C+x%29+=+f%5E%28-1%29%28x%29\"$

\n" ); document.write( "(NOTE: Algebra.com's software mistakenly puts a multiplication symbol in the inverse function notation. This is a mistake. There is no multiplication involved.)
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