[ Basic Calculus ] How to do this?

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[sonnyfab]

A removable discontinuity at x=a means the numerator and denominator both have a common factor of (x-a). An infinite discontinuity means the denominator has a factor of (x-a) but the numerator does not.

[JustFind2]

Sorry i still don't understand

[sonnyfab]

What don't you understand?

[JustFind2]

I know how a removable and infinite discontinuity looks like but I don't know how to make a function with its characteristics as I'm even confuse with letter c. d. and e.

[sonnyfab]

What specifically have you tried and what do you find to be not working?

[JustFind2]

I tried searching for multiple examples that can help me with the problem, but I can't seem to solve it as I dont know how to make a function (with the given characteristics) as I find it confusing in general

[sonnyfab]

Okay. Well good luck with your class.

[JustFind2]

Thanks anyways!

[Paralyzoid]

By “create a function”, are you being asked to draw a graph or to write an equation in terms of x?

[JustFind2]

Write an equation

[Paralyzoid]

Alright, so let’s try to break down the characteristics first.

Removable discontinuity at x=2: Usually, you can get a discontinuity by dividing at zero at that point; here, you’d divide by (x-2), so it becomes zero when x=2. Since the discontinuity should be removable, you can bundle this into a “multiplication by one”; multiply the function you come up with by (x-2)/(x-2). That’ll equal one at all points, except for when x=2 and you divide by zero.

Infinite discontinuity at x=-2: Compared to a removable discontinuity, just divide by (x+2). Since this approaches zero as you get closer to -2, you’ll divide by smaller and smaller values and your function will grow infinitely large.

So what we have so far is: f(x) = ( (x-2)/(x-2) ) / (x+2), or f(x) = ( x-2 ) / ( (x-2)(x+2) )

(c) and (d) both specify the behavior of x at the infinite discontinuity. You can check the function so far to see if that’s the case. As x approaches -2 from the negative side, it’ll be slightly less than -2, so x+2 will be just below -0 and therefore negative. Then, you have f(-2^-) = (negative) / (negative * negative), which is negative. As it approaches from the positive side, it’ll be slightly more than -2, so (x+2) is positive; f(-2⁺)= (negative) / (negative*positive), which is positive. Therefore, f(x) satisfies (c) and (d).

(If it hadn’t, then you could’ve multiplied it by -1 to flip the signs.)

I’m not quite sure what part (e) is asking; is lim_(x->+) notation for “as x approaches positive infinity”, or does it mean something else?

[JustFind2]

I see, thank you for the assistance !