Redefining a function to remove discontinuity?

Question

for example
f(x) = x^2+4x+2/x^2-2x

w redefine it as

g(x) = x^2+4x+2/x^2-2x if x is not = 2
&
g(x) = 5 if x=2



i don't get it.. how can u just redefine a function ??? and if we were asked to redefine a function to make it continuous everywhere what value should we equal it to?

Answer 1

let the function be y = 1 at x = 0.
Your example is not valid because the function does not approach the same finite value either side of the undefined point. It even approaches positive infinity on one side and negative infinity on the other. Just saying let g(2) = 5 makes no sense at all.
The example given in the first answer is not valid either, as it cancels to h(x) = x + 3. There is no 'real' point of discontinuity.


Your original function has something called a "removable discontinuity" a.k.a. a hole in the graph. By redefining it, you are essentially filling in the hole so that it's continuous for all x in the domain.

You CANNOT redefine a function that has an infinite discontinuity, which is a vertical asymptote in the graph of the function.

Answer 2

Your original function has something called a "removable discontinuity" a.k.a. a hole in the graph. By redefining it, you are essentially filling in the hole so that it's continuous for all x in the domain. Here's the method for filling in the hole:

Factor your function and cancel out the common factor from the numerator and denominator (that's the thing that causes the hole in the first place). For example, let's say f(x) = (x^2-9) /( x-3). This factors into f(x) = (x-3)(x+3) / (x-3). The (x-3) cancels, yet x=3 is still undefined in the original function.

So f(x) = (x^2-9) /( x-3) when x is not = 3
= 6 when x - 3.

How did I get the 6? After canceling out the incriminating factor, I had x+3 left. Plug x = 3 into that to get 6. That means if you hadn't redefined the function, the graph would have a hole at (3.6). But when you add on the extra piece to the function that f(x) = 6 when x = 3, you are "filling" in the hole so it's continuous everywhere.

Note: You CANNOT redefine a function that has an infinite discontinuity, which is a vertical asymptote in the graph of the function.

Answer 3

Redefining a function to remove a discontinuity is only really permissible when the function approaches the same finite value from both sides. A good example is y = (sinx)/x. This is undefined at x = 0 but if you draw the graph it looks as though it isn't; it approaches y = 1 from BOTH sides. (The both is crucial as is the fact that it is a finite value.) It therefore seems reasonable to acknowledge this by saying we will let the function be y = 1 at x = 0.
Your example is not valid because the function does not approach the same finite value either side of the undefined point. It even approaches positive infinity on one side and negative infinity on the other. Just saying let g(2) = 5 makes no sense at all.
The example given in the first answer is not valid either, as it cancels to h(x) = x + 3. There is no 'real' point of discontinuity.

Answer 4

Usually in problems like this, the discontinuity is only at a point, that is, there is a hole in the graph. Re-defining the function at that point can make it continuous. That's why in your example, the only point that was added was at x = 2 (because that's where the denominator of the original function was equal to zero).

Here's another example:

h(x) =(x^2 + 6x +9) / (x+3)

is redefined as follows:

h(x) =(x^2 + 6x +9) / (x+3) if x is not = 3
&
h(x) = 6 if x = 3.

I knew the value had to be six because my original function is equivalent to (x+3) (to get this, factor the top, and cancel).

Hope this helps.

Answer 5

When you redefine a function it is no longer the same function as before it is a difrferent function that your using just so you can compute certain things

If the function is undefined anywhere you have to change it. You can define the function as a piece wise function with a certain value at the point of discontinuity.

see the example above and try drawing a graph to understand why they said g(x)=5 at x=2

It will help