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Another useful function is the inverse function. This function undoes the action of the corresponding original function. Inverse functions obey the following laws

 \begin{array}{l}{f^{-1} (f(x))=x \quad{\rm for}\, x\, {\rm in}\, {\rm the}\, {\rm domain}\, {\rm of}\, f} \\{f(f^{-1} (x))=x\quad {\rm for}\, x\, {\rm in}\, {\rm the}\, {\rm domain}\, {\rm of}\, f^{-1} }\end{array}

The following is a procedure for finding the inverse of a function. Suppose you have a function defined as f(x). Compute x=f(y) and then solve for y in terms of x. Then you will have

y=f^{-1} (x)

Example 1. Find the inverse of the following function

f(x)=2x+3

Solution 1. We first write

x=2y+3

Now we solve this in terms of y.

y=\cfrac{x-3}{2}

Now we can read off the inverse

f^{-1}(x)=\cfrac{x-3}{2}

We can check this in the inverse by plugging it into the original function

f(f^{-1}(x))=f\left(\cfrac{x-3}{2} \right)=2\left(\cfrac{x-3}{2} \right)+3=x

f^{-1}(f(x))=\cfrac{(2x+3)-3}{2} =x

It is possible that your function is too complicated to find its inverse, but the inverse exists just that you can’t find a general formula for it. A numerical solution may be necessary.

f(x)=1+x+x^{2} +x^{3} +x^{4} +x^{5}

Good luck finding the inverse of that

 {x=1+y+y^{2} +y^{3} +y^{4} +y^{5} =\cfrac{1-y^{6} }{1-y} }

{y(x)=?}

It is important not to confuse the terms reciprocal and inverse. The reciprocal is one divided by a function $latex [f(x)]^{-1}&s=1$ and the inverse is given when solving by the above method for $latex f^{-1}(x)&s=1$.

Example 2. Find the inverse of $latex f(x) = \frac{1}{1+x}&s=1$ and find the reciprocal also

Solution 2. The reciprocal is easy, just flip the numerator and denominator

[f(x)]^{-1} = 1+x

The inverse is also easy, just solve $latex x= 1/(1+y)&s=1$ which give $latex yx = 1-x&s=1$ or $latex y = \cfrac{1-x}{x}&s=1$

This means the inverse is $latex f^{-1} (x) = x^{-1} -1 &s=1$.

Example 3. Find the reciprocal and inverse of $latex f(x) = x&s=1$

Solution 3. The reciprocal is when you flip the numerator and denominator.

 x^{-1} = \cfrac{1}{x}

Notice that the inverse of $latex f(x) = x&s=1$ can be computed as x = y or $latex f^{-1}(x)=x&s=1$. The function x is its own inverse. For that reason, it is sometimes called the identity function. The reciprocal and inverse are different for this case.

Example 4. Find the reciprocal and inverse of $latex f(x) = \frac{1}{x}&s=1$

Solution 4. The reciprocal is just

[f(x)]^{-1} = \cfrac{1}{f(x)} = x

The inverse of $latex f(x) = 1/x&s=1$ can be found from $latex x = 1/y&s=1$. The inverse of $latex f(x) = 1/x&s=1$ is just $latex f^{-1}(x) = 1/x&s=1$. Don’t get confused and just assume the inverse and reciprocal are just the same for simple expressions like x and 1/x.

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