determine where the function f(x) is continuous (x^2+7)/(x^2+x-6)
The question here is where is it _not_ continuous?
That's at the points where it's undefined, or makes a jump.
This function has undefined points. Can you see why? It's because the denominator is going to be 0 sometimes.
If that is a rational function (a fraction) then we find where the denominator is equal to zero. Whenever the denominator is equal to 0 we have a discontinuity. That is the function is not defined because we cannot divide by the number 0.
so this is not ever going to have the denominator = zero?
So we solve for x^2+x-6=0 we can use the quadratic equation or factor it.
Exactly. Thus, we are looking for when the the denominator is equal to zero and exclude (do not include) this for the domain (allowed values of x).
Can you solve for x^2+x-6=0?
so basically this is never continuous?
Not quite. This function is continuos everywhere except when x^2+x-6=0 (i.e whichever x values satisfy the equation x^2+x-6=0)
-3,2
exactly!!!
:-) awesome
So we conclude: the function f(x) is continuous everywhere except when x=-3 and x=2
hmm, this concept confuses me! what happens to the numerator? it doesn't matter?
this is a better way to look at it. When does a fraction make sense? a/b makes sense whenever b is not 0. a can be any number, since we can divide any number by another (except for 0)
where the function is continuous means for which x-values does our function defined (for which x-values can we evaluate our function)...we cannot evaluate the function when we divide by 0 i.e. when the denominator is 0 b/c we don't know how to divide by 0
***is our function defined (not "does our function defined)
exactly the numerator can equal anything because whichever values of x we plug in we can evaluate it
hmm okay! I hope I can burn this concept into my brain!
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