#
Continuous Functions - Problem 3
*
*2,893 views

The composition of continuous functions is also continuous. So, if f(x) and g(x) are continuous functions, meaning that they are continuous at all points at which they are defined, then f(g(x)) is also continuous. Recall that f(g(x)) means replacing x in f(x) with the function of g(x). For example, if f(x)=x^2+3x and g(x)=2x, then f(g(x))=(2x)^2+3(2x)=4x^2+6x. Notice that, since f(x) and g(x) in this example are both polynomials, they are both continuous. Therefore, their composition, which is also a polynomial, is also continuous. For functions that are not polynomials, before applying the rules of continuous functions (sums, products, quotients, and compositions of continuous functions are also continuous), you must check where the function is defined, and if the function is continuous at each point where it is defined. For example, log and rational functions are not defined over all of the real numbers, so you must first find the points at which they are defined, then check if they are continuous at those points.

So just to review, we have come up with a lot of results about continuous functions. It turns out that if two functions are continuous, then their sum is continuous. Their product's continuous. Their quotient's continuous. A constant times one of the functions also continuous. It turns out that a composition of those two functions will also be continuous. So if two functions f, and g are continuous, then f(g(x)) will be continuous wherever it's defined.

Now using all of these rules, we can determine that pretty much any function that we study is continuous, and where it's continuous. For example, this function where is this function continuous? Well, all we have to do is figure out where it's defined.

One way to do that is, let me rewrite this function by making the multiplication of x plus 4, over x plus 4. This is going to eliminate this little denominator here. Now when I multiply the x plus 4 over the top, I get cancellation which leaves an x plus. I get 3 times x plus 4 which is 3x plus 12.

Then in the denominator, I get x plus 3 times x plus 4. I'm just going to leave that as x plus 3 times x plus 4. In the numerator I can simplify if I like 4x plus 12. X plus 3, x plus 4. When you write it this way, it's clear that this function f(x) is a rational function. Rational functions are continuous functions. So they're continuous wherever they're defined. This one is not defined at -3, or at -4. So f(x), this function, is continuous for x not equal to -3, or -4.

Now let's take a look at this function. G(x) equals 2x minus 50 over root x minus 5. This is not technically a rational function. Rational functions are polynomial divided by another polynomial. This is not a polynomial. However, it is a combination of functions that we know to be continuous. So all we have to do is figure out where this function is not defined, and then we'll know where it's not continuous.

Now this function will be defined as long as this radical is defined, and as long as the denominator is not 0. So it's defined for x greater than or equal to 0. That will keep the radical happy. Root x minus 5 not equal to 0. That'll keep the denominator happy. So root x cannot equal 5. That means x cannot equal just squaring both sides 25. So this function is continuous for x greater than or equal to 0, but not x equal to 25. This is g(x). This is continuous, for x greater than, or equal to 0, but not x equal to 25.

Finally here is a chance for me to use rule 5. Remember that rule about compositions of continuous functions. If I have two functions; f and g that are continuous, their composition is also continuous. Well, here h(x) is a log of a rational function. Rational functions are continuous wherever they're defined. Logs are going to be continuous too. So let's see where this function's defined, and we'll determine where it's continuous.

Now a log is defined as long as the inside part is positive. So we would need 3 minus x over x plus 1 to be positive. Now do you remember how to solve a rational inequality like this? I'll show you really quickly. The way to do it is to set up a sign chart. It's basically a number line. On this number line, you want to look at 3 minus x over x plus 1. You want to analyze for what values of x this quantity is positive, negative or 0.

So look at the zeros of the various factors. We have a factor 3 minus x that has a 0 at 3. This one has a 0 at -1. So you want to put those guys on your sign chart. Now when x equals 3, this quantity is going to be 0. When x equals -1, this quantity is undefined. Now since this function is continuous, it really has no choice, but to be either positive or negative in each of these three regions.

So we test a point in each region, and see whether it is positive or negative. Let's try the left region first. So if I plug in -2 for example. I get 3 minus -2, 3 plus 2, 5. -2 plus 1 is -1. I get 5 over -1. It's going to be negative. I didn't have to know what the answer is. It's going to be negative.

Then let me try something between -1, and 3 like 0. I plug 0 in, I get 3 minus 0 which is 3. 0 plus 1, 1. Positive over positive, this is positive. Finally, let me try a number bigger than 3 like 4. 3 minus 4 is -1. It's negative, over 4 plus 1, 5 positive. So negative over positive is negative.

Now what we have here is that this rational is only positive between -1 and 3, in this interval. That's precisely where this function is defined. That's where this function is continuous. So h(x) is continuous for x between -1 and 3. Remember the combinations of continuous functions are continuous, wherever they're defined.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete