Solving Equations and Finding Inverse Functions
How to solve equations, the systematic way.
To a beginner, there often seems to be no systematic way to solve an equation. It can seem like one just has to "know" what to do before one can do anything.
Here I will show you a technique you can use to solve many types of equations, in an algorithmic fashion.
Let's consider the equation
3(y -2)=x
which we want to solve for y (i.e. we want to end up with something of the form y = "some other stuff").
Here's a longer way of doing it, and below a faster way.
Distribute the 3 out:
3y-6=x
add 6 to both sides:
3y = x+6
and divide by 3:
y=(x+6) = x/3 +2
3
Notice how this approach was inefficient since we multiplied the 3 out but eventually had to divide by 3 again to get y by itself.
Another way to do it is to think of y all by itself, and then to imagine 3(y -2) as some number of things that have happened to y, using the order of operations (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction, aka "PEMDAS").
We start with y
y
we subtract 2
y -2
we multiply by 3
3(y -2).
Let's list these steps in the order in which they occur.
1) subtract 2
2) multiply by 3
Think of y all by itself as a rock about the size of your fist. Then all the other stuff on the y-side of the equation is like a series of papers wrapped around the rock. How we undo the stuff to get y by itself, and how we peel the papers off the rock, is by undoing the later ones first.
So let's undo the last step first, and undo the first step last:
1) subtract 2
2) multiply by 3
inverted becomes
2) divide by 3
1) add 2.
Let's apply this result to the original equation:
3(y -2)=x
divide by 3
y -2=x/3
add 2
y=x/3+2
Notice how this approach was more streamlined, and requires fewer steps, time and energy (one you learn how to use it) than the previous method.
How to find an inverse function
The technique above can be extended to finding inverses to many types of functions.
It works for any function that is formed from compositions of the "simple" functions starting at the identity function f(x)=x.
Let's say you have the function
f(x) = 2[x^3+5]^(1/7)
How could we find the inverse?
Let's think about what happens to x, in order, using order of operations. If we list the steps in order, we get:
^3 +5 ^(1/7) *2
The inverse function will do the opposites of these operations in reverse order, so that we get
/2 ^7 -5 ^(1/3)
This will give us f inverse: [(x/2)^7-5]^(1/3)
Notice that if we put the steps of f inverse after the steps of f, we get
^3 +5 ^(1/7) *2 /2 ^7 -5 ^(1/3)
\ / \ /
f f inverse
Notice how the steps cancel out:
^3 +5 ^(1/7) (*2 /2) ^7 -5 ^(1/3)
^3 +5 ^(1/7) ^7 -5 ^(1/3)
^3 +5 [^(1/7) ^7] -5 ^(1/3)
^3 +5 -5 ^(1/3)
^3 (+5 -5) ^(1/3)
^3 ^(1/3)
[^3 ^(1/3)]
So if we start with f(x) and apply f inverse, we are just left with x. The same thing happens if we start with f inverse (x) and apply f.
A word of caution, be careful when "undoing" a given step. Notice that above we took the cube root in finding f inverse. This is fine because x^3 has an inverse over all real numbers. But let's think about x^2. Would we say the inverse to the x^2 is square root of x, or the negative square root of x? Remember, a function must pass the "vertical line test", so we can't just say "it's plus or minus the square root of x." The square function does not have an inverse, unless we restrict to either positive or negative values.
To a beginner, there often seems to be no systematic way to solve an equation. It can seem like one just has to "know" what to do before one can do anything.
Here I will show you a technique you can use to solve many types of equations, in an algorithmic fashion.
Let's consider the equation
3(y -2)=x
which we want to solve for y (i.e. we want to end up with something of the form y = "some other stuff").
Here's a longer way of doing it, and below a faster way.
Distribute the 3 out:
3y-6=x
add 6 to both sides:
3y = x+6
and divide by 3:
y=(x+6) = x/3 +2
3
Notice how this approach was inefficient since we multiplied the 3 out but eventually had to divide by 3 again to get y by itself.
Another way to do it is to think of y all by itself, and then to imagine 3(y -2) as some number of things that have happened to y, using the order of operations (Parentheses, Exponents, Multiplication, Division, Addition, Subtraction, aka "PEMDAS").
We start with y
y
we subtract 2
y -2
we multiply by 3
3(y -2).
Let's list these steps in the order in which they occur.
1) subtract 2
2) multiply by 3
Think of y all by itself as a rock about the size of your fist. Then all the other stuff on the y-side of the equation is like a series of papers wrapped around the rock. How we undo the stuff to get y by itself, and how we peel the papers off the rock, is by undoing the later ones first.
So let's undo the last step first, and undo the first step last:
1) subtract 2
2) multiply by 3
inverted becomes
2) divide by 3
1) add 2.
Let's apply this result to the original equation:
3(y -2)=x
divide by 3
y -2=x/3
add 2
y=x/3+2
Notice how this approach was more streamlined, and requires fewer steps, time and energy (one you learn how to use it) than the previous method.
How to find an inverse function
The technique above can be extended to finding inverses to many types of functions.
It works for any function that is formed from compositions of the "simple" functions starting at the identity function f(x)=x.
Let's say you have the function
f(x) = 2[x^3+5]^(1/7)
How could we find the inverse?
Let's think about what happens to x, in order, using order of operations. If we list the steps in order, we get:
^3 +5 ^(1/7) *2
The inverse function will do the opposites of these operations in reverse order, so that we get
/2 ^7 -5 ^(1/3)
This will give us f inverse: [(x/2)^7-5]^(1/3)
Notice that if we put the steps of f inverse after the steps of f, we get
^3 +5 ^(1/7) *2 /2 ^7 -5 ^(1/3)
\ / \ /
f f inverse
Notice how the steps cancel out:
^3 +5 ^(1/7) (*2 /2) ^7 -5 ^(1/3)
^3 +5 ^(1/7) ^7 -5 ^(1/3)
^3 +5 [^(1/7) ^7] -5 ^(1/3)
^3 +5 -5 ^(1/3)
^3 (+5 -5) ^(1/3)
^3 ^(1/3)
[^3 ^(1/3)]
So if we start with f(x) and apply f inverse, we are just left with x. The same thing happens if we start with f inverse (x) and apply f.
A word of caution, be careful when "undoing" a given step. Notice that above we took the cube root in finding f inverse. This is fine because x^3 has an inverse over all real numbers. But let's think about x^2. Would we say the inverse to the x^2 is square root of x, or the negative square root of x? Remember, a function must pass the "vertical line test", so we can't just say "it's plus or minus the square root of x." The square function does not have an inverse, unless we restrict to either positive or negative values.