How to find a formula for an inverse function
Remember that $y=f(x)$ means the exact same thing as
$x=f^{1}(y)$. To find a formula for $f^{1}$,
 Solve for $x$ in terms of $y$.
 Now you have $x=$ (a function of y) = $f^{1}(y)$.
 If you wish to have $x$ be the independent variable
(and if the variables $x$ and $y$ have no applicable
meaning), simply exchange $x$ and $y$ to get
$y=f^{1}(x)$.

Example 1: Find a formula for the inverse to $f(x) = 2x+1$.
Solution: Set $y=2x+1$. By definition, $$y =
f(x)\quad \Longleftrightarrow\quad x = f^{1}(y).$$ Therefore, we
have: \begin{eqnarray} 2x+1&=&y \cr 2x&=& y1
\cr x &=& \frac{y1}{2}. \end{eqnarray} So,
$x=f^{1}(y) = \dfrac{y1}{2}$. You would leave the inverse in
this form if, for example, $y$ represents the cost of $x$
widgets. In this case, $y=f(x)$ tells us the cost, $y$, for
any given number of widgets, $x$. $x=f^{1}(y)$ gives the
number of widgets, $x$, for a given cost $y$. If $x$ and $y$
have no meaning, we may wish to rewrite the inverse of $f$ with the
independent variable $x$: $f^{1}(x) = \dfrac{x1}{2}$.
Warning:
in general, $f^{1}(x)$ is not the same as $f(x)^{1}$!
Inverses and reciprocals are different things.
Example 2: Find a formula for an inverse to $f(x) = 2x^2
+ 5$.
Solution: In this case, $f$ does not pass the
horizontal line test, so $f$ does not have an inverse defined on
the whole domain. Let's restrict the domain to $[0,\infty)$, for
example, so that now $f$ does pass the HLT. DO: Why does $f$ now pass the HLT?
The range of $f$ is $[5,\infty)$. Why?
To find a formula for $f^{1}$, we write: \begin{eqnarray}
y&=&f(x) \Leftrightarrow x=f^{1}(y) )\cr 2x^2 + 5
&=&y\cr x^2 &=& (y5)/2 \cr
x&=&\pm\sqrt{\frac{y5}{2}}=f^{1}(y). \end{eqnarray} But
here, $x$ is not a function. So we must determine which
square root we want. Since the range of $f^{1}$ is the
domain of $f$, our answer must be in $[0,\infty)$, so we want the
positive root: $f^{1}(y) = x=+ \sqrt{(y5)/2}$. This makes sense
whenever $y \ge 5$. The domain of $f^{1}$ is $[5,\infty)$, which
is the range of $f$. At this point, if $x$ and $y$ have no
applicable meaning, you could switch $x$ and $y$ if needed to get
$y=f^{1}(x)=\sqrt{\frac{x5}{2}}$.
