<PRE><B>Question 229892</B>
To find the inverse of a function:&lt;ol&gt;&lt;li&gt;If the function is written in function notation replace the f(x) (or whatever) with a y.&lt;/li&gt;&lt;li&gt;Rewrite the equation with x in the place of y and y in the place of x. This changes the equation from the equation of the function to the equation of the inverse relation.&lt;/li&gt;&lt;li&gt;Solve the inverse equation for y, if you can.&lt;/li&gt;&lt;li&gt;If you are able to solve for y and express y as equal to a single expression (without use of +-)&lt;ul&gt;&lt;li&gt;The inverse is a function&lt;/li&gt;&lt;li&gt;The original function was one-to-one. This is so because&lt;ul&gt;&lt;li&gt;The original function, as all function do, maps each x to a single y.&lt;/li&gt;&lt;/li&gt;The inverse which we now have shown to be a function, also maps each of its x&#39;s to a single y.&lt;/li&gt;&lt;li&gt;Since the inverse is the function with its x&#39;s and y&#39;s swapped, an inverse which is a function is mapping each of the y&#39;s of the original function to a single x of the original function.&lt;/li&gt;&lt;li&gt;So each x is mapped to a single y and each y is mapped to a single x. This is what one-to-one means.&lt;/li&gt;&lt;/ul&gt;&lt;/li&gt;&lt;/ul&gt;&lt;/li&gt;&lt;/ol&gt;
Let&#39;s try this on your function:
{{{f(x)= (1/3)x^3+1}}}
1. Replace f(x) with y:
{{{y= (1/3)x^3+1}}}
2. Swap the x&#39;s and y&#39;s. This creates the equation for the inverse relation:
{{{x= (1/3)y^3+1}}}
3. Solve for y if you can:
Subtract 1 from each side:
{{{x-1=(1/3)y^3}}}
Multiply both sides by 3:
{{{3x-3 = y^3}}}
Find the cube root of each side. (Note: If this was an even-numbered root instead of an odd-numbered root, we would have to use a &quot;+-&quot; on the root and the inverse would &lt;b&gt;not&lt;/b&gt; turn out to be a function.)
{{{root(3, 3x-3) = y}}}
4. We were able to solve for y. For each x, there is only one value for {{{root(3, 3x-3)}}}. Our inverse is a function. So f(x) is one-to-one.</PRE>