Relative Extrema, Local Maximum and Minimum, First Derivative Test, Critical Points- Calculus
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- Опубліковано 19 лют 2016
- This calculus video tutorial explains how to find the relative extrema of a function such as the local maximum and minimum values using the first derivative test.
Introduction to Limits: • Calculus 1 - Introduct...
Derivatives - Fast Review:
• Calculus 1 - Derivatives
Introduction to Related Rates:
• Introduction to Relate...
_____________________________
Extreme Value Theorem:
• Extreme Value Theorem
Finding Critical Numbers:
• Finding Critical Numbers
Local Maximum & Minimum:
• Finding Local Maximum ...
Absolute Extrema:
• Finding Absolute Maxim...
Rolle's Theorem:
• Rolle's Theorem
________________________________
Mean Value Theorem:
• Mean Value Theorem
Increasing and Decreasing Functions:
• Increasing and Decreas...
First Derivative Test:
• First Derivative Test
Concavity & Inflection Points:
• Concavity, Inflection ...
Second Derivative Test:
• Second Derivative Test
_________________________________
L'Hopital's Rule:
• L'hopital's rule
Curve Sketching With Derivatives:
• Curve Sketching - Grap...
Newton's Method:
• Newton's Method
Optimization Problems:
• Optimization Problems ...
_______________________________________
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Got a midterm in 10 minutes wish me luck
A midterm in November??
@@IUsePCP in some classes, any test is called a midterm... so if there are like 5 tests throughout the semester, each one is called a midterm even though it's not the middle of the term.
How'd it go? I am going to have my finals tomorrow and I have no idea what this is all about
@@denverdean2663 well, I don't know about him but I had just had a midterm and it went horrible... like really really really bad.
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I really enjoy watching your videos. They have helped me greatly in the last year or so of mathematics.
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If you need to find out if it is a max or min, use this: f’’(x)>0 = min or f’’(x)
This is much easier thanksss T-T
I’m gonna try this out on my test which is in a couple hours. I trust you
if we get a f"(x) = 0 is there still a possibility for x be to be a maxima or a minima if f'(x)=0
is the change in slope (i.e the second derevitive) of a * maxima * or a * minima * must always be * negetive dec. * or
* positive inc. * respectively or can it be * zero * at any of them aswell?
You did a great job with your presentation. Thank you.
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It's crazy how my professor spent 2 hrs and 15 minutes explaining this with a 15 page long worksheet and I didn't understand jack. You literally taught it in 12 minutes, I love you
Watching this 20 minutes before my calc final. Thank you sir
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Thank you starting to feel more confident in BUS CALC
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amazing video. used some great tricks and hacks to get the answers and made everything clear.
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Thank you. I was struggling to understand this . After watching your video I feel much better.
were you sick?
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I’m from Taiwan and I think this video is more useful to me than all the other Chinese calculus video
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After 5 years this video is still so relevant wow great job
yea... the rules of math and calculus don't change much do they...
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idk why or how but this taught me on 5 minutes what I could not figure out from the textbook in over an hour
Thank you . It is amazing for both learning and revising!!!!
9:40 the easiest thing to do is plug in 0. that way, you can cross out all the 'x's and focus only on the signs of the constants. in this case, you have 3 negatives, so you'll get a negative answer after plugging in 0. so the sign between -2 and 1 is '-'. now you can use the multiplicity trick.
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Got a final term in 30 minutes and I don't have any idea what this is🥲 wish me luck
How was it?
He might have failed and gave up. That's why no reply.@@Tayspeaksfacts
This stuff seems complicated at first but after a bit of studying and plenty of practice fits together elegantly.
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there is a mistake at 10:31. (x-1)^2 has even multiplicity, the sign should be + in the number line
that's true cuz i got confused over there
@@hawamusah-opata9799 wouldn't it be "-"? If the power is even, the sign stays the same(like he mentioned in the video). When x is greater then 2, it's negative. Therefore when x is lesser than 2, it's still negative.
I was looking for this comment. Yes I got confused too. I watched it over and over again.
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1:06 here lies the issue with my highschool teacher ;D,
they explain very well everything graphically with simple functions, and the calculations are easy, then we go to functions with critical points that are not extremas, and not defined functions and all that sht and they are like, ey analogically from what we've discussed, like i can imagine graphs as easy as a person who worked this field for 40 years;
it is way easier to remember when you study from raw methods, you cant force intiution in student's brain, it is something that develops over time
We're having midterm exam today so i need to rewatch it again while answering my exam😊😊
bruh
how did it go
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The way he says increasing and decreasing is satisfying for some reason lol
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Good luck lol mine on Friday
Have a Calc test tomorrow
@@foca7550 Same but mine's tomorrow. A day after yours. How'd it go ?
@@21Screen Pretty poorly. I'm in uni and my gf collapsed with severe chest pains the morning of the test. Had to call an ambulance. Was just worried about her the whole time I was taking the test. Good luck on yours though!
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Got my Semester final in 4 hrs. Wish me luck guys.
Got a final in two days wish me luck I’m so stressed
will the critical values be substituted in real function or the marginal function? . If we have only one critical value so are we going to substitute the two extreme values located to the left and side of the critical value? eg if we have the critical value of 4 , the two extreme values located near 4 are 3 and 5 , are we going to substitute these values?
I want to ask, at 7.00 where did the number 6 go? on the question after the factor.
many thx
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My exam is today wish me luck x
At 3:49 when finding the Derivative. Did you factor out the 2 to find the dy/dx?
he factored it just to simplify and make it easier to solve for x and find the critical point. you should always simplify as much as you can (unless its a super complicated problem and it'll just take you back to the original answer)
wow just wow!
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Take note that another way to find whether the point is a maximum or minimum is taking the second derivative and plugging the x value. If the result is x>0 then it is a minimum and if x
Can you find the max and min values of a function with an imaginary critical value? I'm trying to solve a homework problrm but the critical value is the square root of -1 and I'm stuck.
What grade are you?
I got a calc final in the morning wish me luck