Continuity 1b - Polynomial/Rational Functions and The Extreme Value Theorem

[adynizamani]

Thanks for the lecture.

I didn't get one thing though. Not so good at maths so I am sorry if it's too basic.

at 4:10, since a>0 then why did you plot "a" on -ve y-axis/+ve x-axis instead of its opposite (since b<0)?

[MathDoctorBob]

@adynizamani You're welcome, and thanks!  I've written the opposite of what I said. The IVT applies to both cases, so I'll annotate to clear up. Good eyes. - Bob

[adynizamani]

@MathDoctorBob I see, that would be nice. :)

[bathingape915]

your a pretty intimidating guy for a math teacher....not your typical teacher for sure lol

[MathDoctorBob]

@bathingape915 The first day of class is always interesting. But I'm pretty much about getting my students to the end of the semester in one piece.  - Bob

[MultiMoe91]

what if the rational function is Zero by Zero does that make it a point of discontinuity?

[MathDoctorBob]

@MultiMoe91 Thanks for the question. Yes. Continuity at a point requires three conditions: 1) the function is defined at the point, 2) the limit exists at the point (and not infinite), and 3) the value of the function at the point is the same as the limit. If we have 0/0, then the function is not defined at the point.

What 0/0 tells us is that there is a chance the limit may exist. If so, we have a removable discontinuity and we can "fix" the hole in the graph. - Bob

[MultiMoe91]

@MathDoctorBob thanks alot Dr bob

that was very help full.

I understand that for the rational function F(x) = (x^2-3^x-4)/ (x)^(1/2)-2

it is continuous everywhere except at the point 4 since substituting 4 in the fuction will give us 0/0.

so the type of the discountinuity is removable?

and i would appreciate it if you tell me how to Redefine the fuction at 4 to make it continuous everywhere :)

[MathDoctorBob]

@MultiMoe91 You're welcome. Glad to be of help.

To see if it removable or not, we have to find the limit at x=4. 0/0 means we need to do more work. The trick here is to multiply by [sqrt(x) + 2] over itself. When you simplify the function becomes (x+1)(sqrt(x) + 2) by canceling the (x-4) terms. Now if we put x=4, the limit becomes 5 x 4 = 20. So this discontinuity is removable by letting F(4) = 20. (Check for the limit: F(4.01) = (.0501)/(.0025) = 20.05.)

[MathDoctorBob]

@MultiMoe91 (cont.) For something not removable, consider F(x) = 1/x at x=0. The limit does not exist at x=0, and no matter how we try to fit a point at x=0, we cannot connect the ends on each side.  - Bob

[MultiMoe91]

@MathDoctorBob Thank you so much Doctor Bob.

i really appreciate it.

you will always be in my prayers.

[MathDoctorBob]

@MultiMoe91 You're welcome, and thanks for the kind words! Please let me know if you have any other questions. - Bob