Inverse Functions on TI83/84
Copyright © 2001–2023 by Stan Brown, BrownMath.com
Copyright © 2001–2023 by Stan Brown, BrownMath.com
Summary:
After you graph a function on your
TI83/84, you can make a picture of its inverse by using the
DrawInv
command on the DRAW
menu.
For this illustration, let’s use f(x) = √x−2, shown at right. Though you can easily find the inverse of this particular function algebraically, the techniques on this page will work for any function.
I’ve compensated for the rectangular viewing window by setting window margins to 0 to 10 in the x direction and 0 to 6.5 in the y direction. (If you don’t know how to graph a function, please review that procedure.)
The graph of an inverse function is a mirror image of the original through the line y = x, and here’s how to plot that inverse function:
Paste the DrawInv command to your
home screen. 
[2nd PRGM makes DRAW ]
Either cursor down to the 8 and press [ ENTER ], or simply press [8 ]. 
Tell the TI83/84 to find the original function in
Y1 . 
Press [VARS ] [► ] [1 ] [1 ].
(If your function was in a different numbered Y
variable, pick that one instead of Y1 .) 
At this point your screen shows this command:
DrawInv Y1 

Now execute the command.  Press [ENTER ]. 
The result is shown at right.
You know from your algebra work that the inverse of
f(x) = √x−2
is
f ^{−1}(x) = x²+2, x ≥ 0
and the graph confirms that.
Each point on the graph of f(x) has a corresponding point on the graph of f ^{−1}(x). For example, f(2) = 0, so (2,0) is on the original graph. (0,2) is on the graph of f ^{−1}(x), and f ^{−1}(0) = 2.
Unfortunately, all you can do with the inverse is look at it. You can’t trace or do other things. But even that helps you check your work. For instance, you see that the inverse of the sample function appears only in the positive x region. The inverse you calculate algebraically, x²+2, has a domain in both the positive and negative reals, but from drawing the inverse on the TI83/84 you can see that you need to restrict the inverse function’s domain to match the restricted range of the original function.
There’s another way you can check your work. Find the inverse
function first, algebraically, and graph it as Y3
when
you graph the original as Y1
. If you do that,
DrawInv Y1
will exactly overlay the graph of your
algebraic inverse.
Caution: Because the screen resolution is low, two different functions sometimes look the same. This method isn’t an absolute guarantee that your work is correct, but it’s better than no check at all.
Now suppose you have to find f ^{−1}(1.5). (This is equivalent to asking how you can find the x that corresponds to a given y.) Of course you can look at it on the graph and estimate, but your calculator can do a better job of the estimation for you. There are two methods, one on the graph and one on the home screen.
intersect
on GraphRemember that f ^{−1}(1.5) is some value, call it a, such that (1.5,a) is on the graph of f ^{−1}(x), and therefore (a,1.5) is on the graph of f(x). In other words, f ^{−1}(1.5) is the x value on the original graph of f(x) where the y value is 1.5.
Using this idea, to find f ^{−1}(1.5) you can plot y = 1.5 and have your calculator find the point where it intersects the graph of f(x). You don’t need the graph of f ^{−1}(x) for this at all.
The graphs are shown at right, and here’s the procedure.
Select the intersect command. 
[2nd TRACE makes CALC ] [5 ] 
The calculator asks “First curve?”

Simply press [ENTER ] to select the first
curve. 
The calculator then asks “Second curve?”  The cursor may have moved automatically to the other curve. If
not, press [▲ ] or [▼ ] until it
does.
Press [ ENTER ]. 
Finally, the calculator asks for your guess.  Usually you can just press [ENTER ]. But if the function
is very complicated, you can use [◄ ] or
[► ] to move the cursor close to the intersection
point and then press [ENTER ], or type in a number and
press [ENTER ]. 
The result is shown at right: the answer is 4.25.
Why is the answer x and not y? Because you’re trying to find f ^{−1}(1.5), the value of the inverse function of 1.5. But as mentioned above, f ^{−1}(1.5) is the number a such that f(a) = 1.5. In other words, because f(4.25) = 1.5, f ^{−1}(1.5) = 4.25.
Caution: Your calculator gives numerical solutions only. To determine whether 4.25 is the exact answer or just a good approximation, you have to check it in the original function.
solve
on Home ScreenYou can accomplish the same thing on the home screen by using
the solve
function.
Select the solve function from the catalog
because it’s not in a menu. (There’s a Solver
command in the Math menu, but setting it up is a little more
work.) 
Press [2nd 0 makes CATALOG ] [ALPHA 4 makes T ], scroll up to
solve( , and press [ENTER ]. 
The first argument is an expression that you want to equate
to zero. You actually want to equate Y1 to 1.5, which is
the same as equating Y1 −1.5 to 0. 
Press [VARS ] [► ] [1 ] [1 ] −1.5 
The second argument is the variable, x.  Press [, ] [x,T,θ,n ]. 
The last argument is your initial guess. Unless the function is pretty complicated, it doesn’t matter what you enter here as long as it’s in the domain of the function. For example, 0 would be a bad choice for f(x) = √x−2 because f(0) is not a real number. Let’s use 6 as the initial guess.  Enter the initial guess and a close parenthesis
[) ]. 
The screen is shown at right. The answer of 4.25 agrees with the graphical method.
Caution: Again, remember that this is a numerical solution and may not be exact.
Converted HTML 4.01 to HTML5.
Improved formatting of radicals and of f ^{−1}, the inverse function sign. Italicized variable names.
Updates and new info: https://BrownMath.com/ti83/