Algebra: Inverse Functions, Finding Inverses

I'm so sorry this post is a week late. Life got busy. :-)

Now that we understand the definition of an inverse function and how to test them, we can learn how to find them. In this post, I'll share my favorite method of finding the inverse. How do you like to do it?

Step 1: Rewrite with the variables y and x. Use y in place of f(x), g(x), etc.

This is straight-forward. If you are given a function in terms of different variables, such as f(b) and b, change them to y and x. Use y for the dependent variable (usually f(x) or g(x), etc.) and x for the independent variable (usually already x, but may be another letter).

Step 2: Swap the independent and dependent variables.

Swap x and y. Shown below.

Step 3: Solve for y.

Using this new equation, solve for y. Shown below.

Step 4: Replace y with the inverse notation of the original function.

In this case, we replace y with f^(-1)(x) because the original function was f(x). If the original function were, say, g(b), then we would replace x with b, and y with g^(-1)(b). Shown below.

Step 5: Test the inverse.

Using the method we discussed in the previous post, test the inverse you found. Those steps are shown below.

Our answer passes the test of the theorem we learned about, so it is correct.

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Let's try another example. This one is a little more complicated, and will show you why it is important to test your answer at the end!

Steps 1-3: Rewrite in terms of y and x; swap the variables; solve for y.

All three of the first steps are shown below. Notice the last step of solving for y. When you take the square root of a variable, you must include both the positive and negative outside of the square root. Therefore, we actually have two answers for the inverse, this time.

Step 4: Rewrite with inverse notation of the original function.

Step 5: Test the inverse(s).

We must test both of our answers. First, we will test the positive version, shown below.

This answer passes the test. Now, we must test the negative version, shown below.

This answer passes the first portion of the test, but fails the second portion. Remember, both parts of the theorem must hold true in order for the two functions to be considered inverses. Therefore, g^(-1)(x) = - sqrt(x-1) is NOT an inverse of g(x) = x^2 + 1! We simply throw out this answer and only give the positive version, shown below.

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That's how I like to find the inverse of a function. :-) If you have another method you prefer, I'd love to hear all about it in a comment!

That concludes this algebra series on inverse functions. If you have a request for my next series, leave it as a comment on this, or any, post, or send it to me via e-mail (find my e-mail on the Contact page). I'd love to hear from you!

Algebra: Inverse Functions, Testing Inverses

Remember that theorem I gave you in the Introductory post for this series?

We can use this theorem to test whether two functions are inverses of each other.

To test whether these two functions are inverses, we need to look at the two composite functions in the theorem above.

Step 1: Evaluate the first composite function in the theorem above, (f o g)(x).

To form a composite function, locate the variable (x) in the first function (circled in red in the image above). Insert the entire second function into this variable (circled in green, above, with arrow showing the insertion point).

Now, evaluate this new function, shown below.

If the first composite function equals x, continue to Step 2.

Otherwise, the two functions are NOT inverses.

In this example, the first composite function, (f o g)(x), does equal x. Therefore, we continue to Step 2.

Step 2: Evaluate the second composite function in the theorem above, (g o f)(x).

Form this composite function as before. This time, the entire first function (circled in red above, with arrow showing insertion point) is inserted into the variable placeholder of the second function (circled in green).

As before, evaluate this new function, shown below.

If the second composite function equals x, the functions ARE inverses.

If EITHER of the two composite functions do not equal x, the functions ARE NOT inverses!

In this example, the two functions are inverses of each other.

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Here's another example.

Begin with Step 1 as before.

The first composite function cannot be reduced any further than (3x^2 - 1), as shown in the image above. Therefore, it does not equal x. And therefore, the functions f(x) and g(x) are NOT inverses.

There is no need to proceed to Step 2. According to the theorem, both composite functions must equal x for the two functions to be inverses of each other.

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The composite functions need not be evaluated in the order given. You could evaluate the second composite function first. Sometimes, one of the two is easier to reduce than the other. If you get stuck trying to reduce one of them, try the other.

In the next, and final, post of this series, we'll discover how to find the inverse of a given function. :-) I hope you'll stick around for it!

Algebra: Inverse Functions, Introduction

Think of a function as a machine. It takes an input (x-value, independent variable), performs some operation (equation) on it, then spits out an output (y-value, dependent variable).

The inverse of a function is the undo button in the world of functions. It takes an output, performs an operation, and gives back the input that would have went into the original function machine.

There is a way to test whether two functions are the inverses of each other. It is actually a mathematical theorem, which says:

We will use this information in the next two posts, which make up this series on inverse functions. In the first, we will discuss how to use the above theorem to test whether two functions are the inverses of each other. In the second, we'll see how to find the inverse of a given function.

I hope you'll stay tuned!

Algebra: Multiplying Polynomials, Method 3

If you missed the introductory post to this series, you should go read it.

This is the last method I'll be sharing for how to multiply polynomials. I call this one the "Numerical" method because it is set-up like a standard multiplication problem with numbers. I do not find this method particularly quick (like "The Table" method), but it is organized and easy to understand because it builds upon a skill with which the majority of students are already comfortable.

Our first example is an easy one, as usual. :-)

Step 1: Write the polynomials in a column, aligning them at the right, with like terms stack (introduce missing terms with a zero coefficient), like you would a standard multiplication problem. Write the longer polynomial on top.

In this example, there are no missing terms. But, if there were, I'd introduce them (as in previous methods), with zero coefficients. Like terms should be aligned. Notice in the image above that the term (-2x) in the first polynomial is aligned in the same "column" as the (+x) in the second polynomial.

Step 2: Multiply the last term of the bottom polynomial by each term of the top polynomial, from right to left. Write these products below the solid line, aligned with the term from the top polynomial. This is just like a standard multiplication problem with numbers.

In this step, we are working with the last term of the bottom polynomial (circled in red in the image above). Make sure you include the sign along with the term when multiplying! First, this term is multiplied by the last term of the top polynomial (circled in green). The product is written below the term used from the top polynomial (underlined in green).

Continue multiplying the last term (red circle) of the bottom polynomial by each of the terms in the top polynomial, from right to left. These are shown in purple and blue, respectively.

Step 3: Multiply the next to last term of the bottom polynomial by each term of the top polynomial, from right to left. Write these products below the products from Step 2, aligning them one term from the right side (leave a space below the far-right term of the first row of products). This is just like a standard multiplication problem with numbers.

Continue in this same manner until every term of the bottom polynomial has been used. The number of rows in the "products" section (below the solid line) should be equal to the number of terms in the bottom polynomial.

Now, we are multiplying across the top polynomial using the second to last term of the bottom polynomial (circled in red in the image above). These products are aligned below the products from Step 2, but moved one "space" to the left. This ensures that like terms are aligned in the products! This should look very familiar to you as a standard multiplication problem.

In this example, the multiplying is complete. Every term of the bottom polynomial has been multiplied through the top polynomial. Notice, there are two rows of product polynomials below the solid line, which is equal to the number of terms in the bottom polynomial (2).

Step 4: Draw another solid line below the rows of product polynomials. Add these polynomials together. Like terms are added together. The final product polynomial is yielded. (If there are any terms with a zero coefficient, rewrite the product polynomial and remove them.)

This step is where like terms are combined to form the final product polynomial. Simply add the terms in each column (example, boxed in green, above). The final product polynomial is boxed in red.

This method can be difficult to explain, but it really does work exactly like multiplying numbers with multiple digits that is taught in elementary/primary school.

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And now for an example with missing terms and longer polynomials. :-)

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I hope you have found this series helpful. As always, if you have any questions, comments, or requests for future topics, leave a comment below. :-)

Algebra: Multiplying Polynomials, Method 2

Read the introduction to this series here.

This second method is my favorite way to multiply polynomials. I call it "The Table" method or "The Box" method. Even way up in Calculus III in university, my homework and test pages would have little boxes in the margins wherever I had to multiply polynomials. Many students find this method easier to follow and more organized. While tutoring, my students rarely missed terms when using this method. However, tables can be confusing in their own right, so this method may not be for everyone!

Let's begin with a simple example.

Step 1: Set up a table with one polynomial across the top (one term per column) and the second polynomial down the left side (one term per row) (including all +/- signs!).

***Important! Introduce any missing terms with zero coefficients (in red on first polynomial, below). Doing this will ensure your like terms align properly when you multiply. With enough practice, you will be able to skip this, but it is very helpful when you first begin using this method!***

In the image below, you can see I have set up my table with the first polynomial across the top and the second down the left-hand side. I included the missing term of the first polynomial, giving it a coefficient of zero (in red).

Step 2: For each cell in the table, multiply the term for that column by the term for that row.

So, begin by multiplying x times x^2, which yields x^3. This cell is marked with a green circle. Continue multiplying in this way until every cell of the table is filled, as shown in the image below. Watch the signs on each term carefully!

Step 3: Beginning with the first cell, write out the product polynomial by combining like terms on the diagonal as you go.

Notice that like terms are collected on diagonals, illustrated with the green and yellow highlights in the image below.

As you write out your answer (boxed in red above), simply add/subtract like terms along each diagonal, working from the top left corner to the bottom right.

You can help ensure you do not miss terms by marking off the cells you have already written as you go.

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Here is another, slightly more complicated example, worked out in steps below:

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As you can see, this method is very compact, even with large polynomials.

There is one more method I'll be sharing with you. Look for it Wednesday, March 18th!

Algebra: Multiplying Polynomials, Method 1

Read the introduction to this series here.

I call this first method "Term-by-Term." While tutoring, I found that most students attempted to multiply polynomials in this way. When the polynomials are small (two binomials, for example), this method is fast and easy to do. However, when the polynomials are large, it is cumbersome and lengthy.

Step 1: Multiply the first term of the first polynomial by each term in the second polynomial (including +/- signs!). 

In the image below, first the operation shown by the red arrow is performed, then the operation shown by the blue arrow. The resulting products (underlined in red and blue) are written out as terms in the product polynomial.

Step 2: Multiply the second term of the first polynomial by each term in the second polynomial (including +/- signs!).

In the image below, the two operations are shown in green and purple. These products, underlined, are tacked onto the products from the previous step in the product polynomial.

Continue multiplying in this way until every term from the first polynomial has been used.

Last Step: Combine like terms and rewrite in standard polynomial format (with exponents in descending order).

In this example, there are no like terms to combine, so we simply need to write the terms in the correct order (shown boxed in red below).

You see, when multiplying two binomials, this method works very well. Some students may find it difficult to keep track of which terms they are currently multiplying. My suggestion is to try drawing the arrows, as I've done, while you multiply one term across the second polynomial. Then, erase those arrows and draw new ones for each successive term.

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Let's try multiplying two trinomials with this method...

The images below show each step.

And the final answer...

Hopefully, this example illustrated how cumbersome this method becomes with long polynomials. Each step takes up more and more space on the page. In the end, there can be dozens of terms to sort through and combine.

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The method next up in this series is much more conducive to multiplying longer polynomials. Look for it next Monday, March 16!

Algebra: Multiplying Polynomials, Introduction

Many of the students I tutored were confused any time they had to multiply polynomials. Over the few years I tutored, I developed three methods for performing this arithmetic. It never failed that at least one of these three methods would "click" with someone who previously struggled to understand how this works.

I'll be sharing these three methods with you in a series of posts, in order to give each method enough attention. My names for the methods are as follows:

• Term-by-Term
• The Table
• Numerical

My personal favorite is "The Table" method -- it is most intuitive to me. Even up through Calculus III in university, my homework and tests would have a little table scrawled in the margins where ever I was required to multiply polynomials. :-)

I hope you find at least one of these methods helpful. I will also give a brief description of when each method may be preferred over another.

Please stick with me through this series. Let me know if you have any questions or comments, or if you have a request for a future post.

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Algebra: Dividing Polynomials, Part 2 of 2

In Part 1 of this series, I explained standard long division of numbers, and we covered a very basic polynomial division problem. Today, we will cover three situations you may encounter: (1) when there is a "missing term" in the dividend, (2) when the leading co-efficient of the dividend is not 1, (3) when there is a remainder.
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Situation #1: How to set up the problem when there is a "missing term" in the dividend.

Look at the problem above. Do you see the where the "missing term" should be? If you look at the exponents of the dividend, you'll see that there is no term for x^2. In order to properly divide these polynomials, we need to reintroduce the missing term. We can do that without changing the value of the dividend by using a co-efficient of zero, as shown circled in red below.

Now that the problem is set up in a familiar way, you can solve it as we did the problem in the Part 1. The steps are shown in the image above.

You should be aware that there may be multiple "missing terms", as in the polynomial (x^5 - x^2 + x). In such a case, introduce enough terms that all exponents are covered. In this example, you would write (x^5 + 0x^4 + 0x^3 - x^2 + x + 0) under the division bar.
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Situation #2: The dividend's leading co-efficient is not 1.

The leading co-efficient in a polynomial is the co-efficient on the term with the highest exponent. In the polynomial (2x^3 - x^2 - 3x + 2), the leading co-efficient is 2 because it is on the term with the highest exponent, 3. All of the problems we have dealt with so far have had a leading co-efficient of one. Let's use the polynomial I've given you already to work another division problem.

The first step of this problem is just like the others we have solved. Begin by dividing the first term of the dividend (green circle) by the first term of the divisor (orange box) to get the first term of the quotient (red box). Then, multiply this first term of the quotient by the whole divisor (underlined in purple). This new polynomial (underlined in blue) is written below the dividend and subtracted. Do you see how this step is just like the first step of the other problems we solved? The only, very minor difference, is the initial division of the first terms of the dividend and divisor.

Now, you can finish solving the problem as you would any other, shown below.

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Situation #3: The remainder is not zero.

Consider the problem I have worked below. You will notice that at the end, I am left with (-5x+1) (red box). I cannot divide (-5x) by the first term of the divisor, (x^2) because (x^2) is larger. What do we do with this remainder?

Your instructor should explain how to represent this remainder. Two representations are given below, although the second looks more professional and is my preference.

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I hope this series has been helpful to you. If you have any questions about this topic or something you would like me to cover in the future, leave them below. Thanks!