Calculus 4.5 An Algorithm for Curve Sketching

A final word on curve sketching where I review all of the key elements of finding the characteristics of a function; s and y intercepts, domain, vertical asymptotes, horizontal asymptotes, using the first derivative to find critical values, the first derivative test to determine if you have a minimum or maximum value, second derivative test as well, increasing and decreasing intervals. Using the second derivative to find points of inflection (if they exist) and also using the 2nd derivative to determine the concavity of the function.

[@gurpreetkaur4602]

It's too great having you! Thanks a lot !!!

[@mshavrotscanadianuniversit6234]

It's my pleasure

[@mallickqamarraza3659]

Nice and effective lesson, it's helpful for me....

[@mshavrotscanadianuniversit6234]

Thank you! 😃

[@Joe-vw7sr]

I just wanted to say thank you!

[@mshavrotscanadianuniversit6234]

You're very welcome! Thanks for taking the time to send me a message. They make my day brighter!

[@plumwine6802]

thank you miss you made it sooooo clear!!!!Love you❤️

[@mshavrotscanadianuniversit6234]

So happy to help you out. Math is tough without good explanations.

[@uzi2neat939]

Hello ms.Havrot thats for the vid. I had a quick question, when finding critical numbers should I first check for when the function is equal to zero or when the function is undefined? or Both?

[@mshavrotscanadianuniversit6234]

I’m sure you meant set the first derivative equal to zero to find where the slope is zero. Also you do need to find where the original function is undefined. So, both! 😊

[@uzi2neat939]

@@mshavrotscanadianuniversit6234 ah okay. I like to make a sign chart with all my critical points and POIs. Should i include my Vertical asymptotes too?

[@Pilaspilas911]

Really helpful. Thank you :)

[@kevinxie1837]

Hello Ms Havrot, in the function y = x raised to the power of 1/2, is the point (0,0) a point of inflection? Thanks!

[@mshavrotscanadianuniversit6234]

No it is not. The function you are asking about is the square root function. At x =0 this is the starting point of the function and to have a point of inflection requires a change in concavity at that point.

[@amanvarghese3937]

Thank you Ms. Havrot. This unit is really tedious with all this writing 😅

[@mshavrotscanadianuniversit6234]

Yeah … there’s a lot to cover. 😳

[@maziam8834]

Thank you so much Ms! What would the range be for this function?

[@mshavrotscanadianuniversit6234]

y is an element of real numbers y> 1, y

[@Gabriel-vk7nt]

this vid a banger

[@mshavrotscanadianuniversit6234]

😂😂❤️ please like,share and subscribe to help increase the channel’s visibility 😊

[@Marvin-us6tm]

All superheroes don't wear capes!

[@mshavrotscanadianuniversit6234]

Wait, I DO wear a cape 😂

[@yuhyuh5674]

Wouldn't the interval of increase be from (-infinity, - 2) U (2, infinity)? It's decreasing from (-2,2)

[@mshavrotscanadianuniversit6234]

Make sure that you are reading from left to right. From -2 to 0 the function is increasing because as you go from left to right the y values are increasing. Does that help?

[@yuhyuh5674]

@@mshavrotscanadianuniversit6234 most certainly. So if I had f(x) = x^2 / x-1, would the function be decreasing from negative infinity to the V. A at 1 and increasing from 1 to infinity?

[@mshavrotscanadianuniversit6234]

This function would have an oblique asymptote because the degree in the numerator is one greater than the denominator. If your teacher hasn’t suggested this site to you you should check it out : www.desmos.com/calculator/ruo0jeaion

[@musamalik6174]

quick question why is there no POI

[@mshavrotscanadianuniversit6234]

There is no point of inflection because when you set the numerator of the second derivative equal to zero you would have x squared equal to a Negative number and that is impossible.

[@bothainaaleleiwi9398]

I love you ❤👍👍

[@mshavrotscanadianuniversit6234]

Awww that’s so nice of you to say. Love you back ❤️Keep watching! 😊