Tuesday 4 March 2014

Posted on March 4, 2014 by Sybil Shaver

Topics:

• Definition of the inverse function

• Finding a formula for the inverse of a one-to-one function algebraically

 

Important Notes:

• For a one-to-one function f(x), the inverse function is denoted by f^{-1}(x). This is read “f inverse of x”. It is not the -1 power! We happen to use the same notation, but when applied to the name of a function it means “inverse” and is not an exponent or power.

• The inverse function reverses the roles of the input and output. Therefore, the domain of f(x) will be the range of f^{-1}(x) and vice-versa.

• To find the formula of the inverse function, we change x to y and y to x in the formula for the original function, and then solve for y. If we can solve for y (the new y) uniquely in terms of x, then the original function was one-to-one and all is well. If we cannot solve uniquely for y in terms of x, it means that the original function was not one-to-one and so does not have an inverse.

 

For example, take f(x) = x^2 (which we already know is not one-to-one). If we go ahead and try to find a formula for its inverse, we get this:

y=x^2 (the original function’s formula)

x=y^2 (reversing the roles of x and y)

y^2 = x (I put the term with y on the left, to make it look nicer)

y = \pm\sqrt{x} (using the square root principle to solve for y)

So we cannot solve uniquely for y in terms of x after reversing the input and output. This is a reflection of the fact that the original function was not one-to-one.

 

• As in the example we worked in class, the inverse function will do the opposite (inverse) operations in the reverse order compared to what the original function does.

 

• The domain of f(x) is the range of f^{-1}(x), and the range of f(x) is the domain of f^{-1}(x).

 

The “round-trip theorem”:

If f and g are one-to-one functions, then they are inverse to each other if and only if both of the following are true:

f(g(x)) = x for all x in the domain of g

AND

g(f(x)) = x for all x in the domain of f.

This theorem provides a way of checking if two functions are inverse to each other.

• The relationship between the graphs: the graph of f^{-1}(x) is the reflection of the graph of f(x) over the line y=x. (This is because they reverse the roles of y and x.)

• For a function which is not one-to-one: Restricting the domain to an interval where the function is one-to-one, so that we can find an inverse on the restricted domain. Note that when we restrict the domain, we want to make sure that the entire range of the function is covered in the restricted domain.

One example we discussed is f(x)=x^2. Here is a bit more detail than I gave in class.

If we restrict the domain to the interval [0, \infty), then f(x) is one-to-one on that interval and every value in the range of f is reached for some value of x in the restricted domain. [This would not be true if, for example, we restricted to the open interval (0, \infty). Then f would be one-to-one on the restricted domain, but the value 0 in the range of f would not be in the new range of the restricted f, because no x-value in the new domain gives f(x)=0.]

 

On that restricted domain, the inverse function is f^{-1}(x) = \sqrt{x}. We can see this by the algebraic method:

y=x^2 (remember that we are taking x \ge 0 when we restrict the domain)

Reverse x and y: x=y^2 (and now y \ge 0)

Solve for y: y^2=x

y=\sqrt{x} – We only take the positive branch of the square root, because y \ge 0 due to the restriction on the domain of f(x). So there is a uniquely defined inverse function for f(x) on this restricted domain.

 

What if we restricted the domain to the interval (-\infty, 0] ? On this interval f is one-to-one (this is just the left-hand side of the graph) and the entire range of f is covered. But in this case, the inverse function on the restricted domain would be f^{-1}(x) = -\sqrt{x}. Why is this so? Because when you try to find the inverse algebraically,

 

y=x^2 (remember that we are taking x \le 0 when we restrict the domain)

Reverse x and y: x=y^2 (and now y \le 0)

Solve for y: y^2=x

y=-\sqrt{x} – We now only take the negative branch of the square root, because y \le 0 due to the restriction on the domain of f(x). So there is a uniquely defined inverse function for f(x) on this restricted domain, but it is different from the inverse we get if we restrict the domain as we did the first time.

 

Moral of the story: there may be more than one choice for how to restrict the domain of a function which is not one-to-one, but different choices will lead to different inverse functions.

• Polynomials: definitions and important vocabulary you should know. (See the list below)

• Long division of polynomials.

Next time we will learn a shortcut efficient method for doing long division of polynomials in certain cases. But there are still times when we will need to use the full long division, so make sure that you practice it!

 

Vocabulary you need to know for polynomials:

• monomial

• polynomial

• term

• coefficient

• degree (of a term, or of a polynomial)

• constant term – note that the degree of a constant term is defined to be 0.

• leading term and leading coefficient

• root of a polynomial (= zero of the polynomial function = x-intercept of the graph of the polynomial)

• Related to division: dividend, divisor, quotient, remainder

 

Homework:

• Reread and review the definitions and examples discussed in class. Make sure that you know and can use all the vocabulary correctly!

• Do the assigned parts of Exercises 7.2-7.4, and 8.1

• Take a look at the “Review of functions and graphs” which follows Session 7. (But skip #1.7, which we did not discuss, unless you want to try it – it’s not hard.)

• Do the WeBWorK: due by Wednesday 11 PM extended to Monday 11 PM. Start early! and use the Piazza discussion board if you get stuck on any of them (or on the regular homework)!

• Read ahead in Session 8 and then do the Warm-Up for Remainders – due also by 11 PM Wednesday  extended to Monday. (This one uses a Google Docs form instead of Piazza. Please only submit your answer ONE TIME. You should print or save the receipt that you get when you have submitted it.)

 

Please note that from now on you are expected to do all the online assignments on time and to inform me promptly if there is any difficulty doing them. I have been very flexible so far, but it’s time to get serious.  I believe we have had time to shake out all the bugs in the system. There will be no more re-assignments or extensions except under very unusual circumstances. So get to it!

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