College Math Teaching

August 20, 2018

Algebra for Calculus I: equations and inequalities

Filed under: basic algebra, calculus, pedagogy — collegemathteaching @ 9:24 pm

It seems simple enough: solve 3x+ 4 = 7 or \frac{2}{x-5} \leq 3 .

So what do we tell our students to do? We might say things like “with an equation we must do the same thing to both sides of the equation (other than multiply both sides by zero)” and with an inequality, “we have to remember to reverse the inequality if we, say, multiply both sides by a negative number or if we take the reciprocal”.

And, of course, we need to check afterwards to see if we haven’t improperly expanded the solution set.

But what is really going on? A moment’s thought will reveal that what we are doing is applying the appropriate function to both sides of the equation/inequality.

And, depending on what we are doing, we want to ensure that the function that we are applying is one-to-one and taking note if the function is increasing or decreasing in the event we are solving an inequality.

Example: x + \sqrt{x+2} = 4 Now the standard way is to subtract x from both sides (which is a one to one function..subtract constant number) which yields \sqrt{x+2} = 4-x . Now we might say “square both sides” to obtain x+2 = 16-8x+x^2 \rightarrow x^2-9x+ 14 = 0 \rightarrow (x-7)(x-2) = 0 but only x = 2 works. But the function that does that, the “squaring function”, is NOT one to one. Think of it this way: if we have x = y and we then square both sides we now have x^2 = y^2 which has the original solution x = y and x = -y . So in our example, the extraneous solution occurs because (\sqrt{7+2})^2 = (4-7)^2 but \sqrt{7+2} \neq -3 .

If you want to have more fun, try a function that isn’t even close to being one to one; e. g. solve x + \frac{1}{4} =\frac{1}{2} by taking the sine of both sides. 🙂

(yes, I know, NO ONE would want to do that).

As far as inequalities: the idea is to remember that if one applies a one-to-one function on both sides, one should note if the function is increasing or decreasing.

Example: 2 \geq e^{-x} \rightarrow ln(2) \geq -x \rightarrow ln(\frac{1}{2}) \leq x . We did the switch when the function that we applied (f(x) = -x was decreasing.)

Example: solving |x+9| \geq 8 requires that we use the conditional definition for absolute value and reconcile our two answers: x+ 9 \geq 8 and -x-9 \geq 8 which leads to the union of x \geq -1 or x \leq -17

The fun starts when the function that we apply is neither decreasing nor increasing. Example: sin(x) \geq \frac{1}{2} Needless to say, the arcsin(x) function, by itself, is inadequate without adjusting for periodicity.

Advertisements

1 Comment »

  1. […] more advanced discussion can be found here; basically, it boils down to: when you do the same process to both sides of an equation, be careful […]

    Pingback by Notes on equations and inequalities | Bradley University Fall 2018 Calculus I — August 20, 2018 @ 9:49 pm


RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

Create a free website or blog at WordPress.com.

%d bloggers like this: