I can't even begin telling you how these videos turned Calculus 1 around for me. My first test was a 60 which was the wake up call. Since then I've found these videos and haven't scored below a 90 on each exam.
This was so helpful. Thank you! My prof just rushes through things, never explains what or why he does things and just assumes we know how to simplify EVERYTHING. I wish he was this clear!
I think the most impressive thing here is how he memorized the mirrored function and uses those hand gestures to make it more clear for the students. If you think about it from his side its really impressive how hes so fluid with it. He has to flip the y axis on any graph he uses a hand gesture for.
I shall quit physical school and just watch Superman's videos. I hate math righ now, but when I understand it I love it. LOVE HATE relationship. Cal1 is not my friend and neither is cal 2, 3 and dif Q. I look forward to taking those classes -______-
### Calculus 1 Lesson 3.3: The First Derivative Test for Increasing and Decreasing Functions Introduction ● [0:00] Professor Leonard introduces the topic: the first derivative test to determine intervals of increase and decrease of a function. ○ Mention of the application of this concept in curve sketching and understanding the shape of curves. Review of Previous Concepts ● [0:55] Review of the meaning of the first derivative and its relationship with a function’s increase and decrease. ○ [1:10] If 𝒇'(x) > 0, the function 𝒇(x) is increasing. ○ [2:16] If 𝒇'(x) < 0, the function 𝒇(x) is decreasing. ○ [2:33] If 𝒇'(x) = 0, the function 𝒇(x) has a horizontal tangent, and a critical number is identified. The First Derivative Test ● [3:05] Introduction to the first derivative test to find relative extrema (maximums and minimums). ● [3:26] Step-by-step explanation of the first derivative test: ○ 1.[4:21] Find the first derivative, 𝒇'(x). ○ 2.[4:40] Set the first derivative equal to zero, 𝒇'(x) = 0, to find critical numbers. ■ Besides the values where 𝒇'(x) = 0, the following points should also be included as “critical numbers,” where 𝒇'(x) does not exist: ◆ Points where the denominator is zero (possible vertical asymptotes). ► Occurs when the function is not defined (or tends to ±∞) as x approaches a certain value x = a. An example is 𝒇(x) = 1/x. As x approaches 0, 𝒇(x) tends to ±∞, and the line x = 0 is a vertical asymptote. Here, the function does not even reach a finite value at x = 0; it simply blows up to infinity (or negative infinity). ◆ Sharp corners or cusps (abrupt changes in slope). ◆ Vertical tangents (infinite slope). ► Occurs when the function is (or can be) defined at that point, but its slope (derivative) becomes infinite (or tends to ±∞). A typical example is 𝒇(x) = x^(1/3) at x = 0. The function is indeed defined at x = 0, but its derivative 𝒇'(x) = 1 / (3x^(2/3)) tends to ±∞ as x approaches 0, indicating a vertical tangent. ◆ Discontinuities (jump, infinite, removable). ○ 3. [4:58] Create a first derivative chart with critical numbers in order. ■ Draw a number line and mark the critical numbers. ■ Place 𝒇'(x) above the number line. ○ 4. [6:16] Test a point in each interval defined by the critical numbers. ■ Determine the sign of 𝒇'(x) in each interval. ◆ Substitute a value from each interval into 𝒇'(x) to determine if the slope is positive or negative. ■ The sign of 𝒇'(x) indicates whether the function is increasing or decreasing in each interval. ◆ Local Maximum: ► If 𝒇'(x) changes from positive to negative at a critical number, there is a local maximum. ◆ Local Minimum: ► If 𝒇'(x) changes from negative to positive at a critical number, there is a local minimum. ◆ Neither Maximum nor Minimum: ► If 𝒇'(x) does not change sign at a critical number, it is not classified as a maximum or minimum. △ Sometimes referred to as a "horizontal inflection point" if: ▢ 𝒇'(x) = 0. ▢ The concavity changes (related to the second derivative). Example 1: 𝒇(x) = x³ -3x +1 ● [10:44] Application of the first derivative test to an example. ○ Find the derivative of the function. ○ Factor the derivative to find critical numbers. ○ Check for values that make the derivative undefined (denominators). ○ Create a first derivative chart with critical numbers. ○ Test a value from each interval in 𝒇'(x) to determine the sign. ○ Interpret the results to determine intervals of increase and decrease. ○ [14:13] Identify relative extrema (maximums and minimums). ○ Find the points of the relative extrema by substituting x-values into the original function. Example 2: 𝒇(x) = 3x^(5/3) - 15x^(2/3) ● [17:12] Presentation of a second example to apply the first derivative test. ○ Find the first derivative of the function. ○ Factor the derivative to find critical numbers. ○ Check for values that make the derivative undefined (denominators). ○ Include values that make the denominator zero in the first derivative chart. ○ Create the first derivative chart with critical numbers, including values that make the denominator zero. ○ Test a value from each interval in 𝒇'(x) to determine the sign. ○ [23:05] Interpret the results to determine intervals of increase and decrease. ○ Identify relative extrema (maximums and minimums), considering values that make the denominator zero. ○ Determine whether the relative extrema are maximums or minimums.
at 4:21 function must be continuous at critical points (where f' is zero or not defined). Your rule is true only for differentiable functions. You restrict the class of functions.
If a function is differentiable, the function MUST be continuous. Existing critical points means that the function has a derivative or is differentiable. Eventually, you are not true.
Starting around 17 minutes...I have a problem with him assigning a positive value to points with negative values under a radical. Seems like positive/imaginary for anything less than 0
@@actuary32174 idk if I misunderstood you or not but if you mean the radical in the second question its odd radical so it's legal to put negatives there
Try substituting -8. It's a cubed root not a square root, so the cubed root of -8 is -2. (Check that (-2)^3=-8). When you plug -8 into the numerator, you also get a negative answer. A negative numerator and a negative denominator give a positive final answer.
for the function f(x) = 3x^(5/3) - 5x^(2/3): We determined in your video that the slope is undefined at x=0. However the original function is continuous at x=0. What does this mean?
If you look at the graph of the function, you can see that the slope is becoming vertical as we approach a sharp turn at x=0 . The slope of a vertical line is undefined, so that is why the first derivative is undefined at x=0 even though the function is continuous there. (The sharp turn also makes the derivative not exist at that point, this function just happens to have both a sharp turn and a vertical slope at the same point)
Functions are allowed to have sharp turns. The absolute value function f(x)=|x| is another example of a function that has a sharp turn at x=0. For the function in the lecture, there is a sharp turn at x=0 but the function is still continuous there. However if you look at the graph you can see that the slope gets steeper and steeper as we approach x=0. The slope becomes vertical, and the slope of a vertical line is undefined. Further, the slope is approaching positive infinity as we approach x=0 from the left, and the slope approaches negative infinity as we approach from the right. So even though the function is continuous at x=0, we see that the first derivative is undefined there so the function is not differentiable at x=0. That doesn't stop us from finding the derivative at the other points where it is defined.
here where Algebra is crutia and will not play with u! it you miss your critical numbers your answer will be in a critical condition and you will find yourself trying to catch up some points at the hospitals problems :D
f(x)=4x+2/2x-3 (4x plus 2, over 2x minus 3). The (1st derivatives) f'(x)= -16/(2x-3)^2 (negative 16 over 2x minus 3 raise to the power of 2), and the f"(x)= 64/(2x-3)^3 When ask when is f'(x) is increasing or decreasing. How to compute when the numerator has no x value). In (second derivatives) f"(x), how to compute when f(x) is positive(conveks) or when it is negative (concave)? Can someone at least explain the logic behind it? Eventually youtube link if there is any.
I think so, yes. I studied pretty much this exact content when I studied "Differential Calculus" on my first year in Engineering Physics. Then my final math course that year was "Multivariable Calculus", which is basically Calculus 3.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
Man is both mentally and physically strong. That's someone to look up to. Great lesson and explanation
"no.. oh no, I'd burn your paper for that... hand ya back a staple!"
This dude is too much hahaha
This guys amazing. Wherever he is teaching, they better be paying him good !
Jbh
I can't even begin telling you how these videos turned Calculus 1 around for me. My first test was a 60 which was the wake up call. Since then I've found these videos and haven't scored below a 90 on each exam.
This was so helpful. Thank you! My prof just rushes through things, never explains what or why he does things and just assumes we know how to simplify EVERYTHING. I wish he was this clear!
I think the most impressive thing here is how he memorized the mirrored function and uses those hand gestures to make it more clear for the students. If you think about it from his side its really impressive how hes so fluid with it. He has to flip the y axis on any graph he uses a hand gesture for.
Professor Leonard is really out here doing God's work
I shall quit physical school and just watch Superman's videos. I hate math righ now, but when I understand it I love it. LOVE HATE relationship. Cal1 is not my friend and neither is cal 2, 3 and dif Q. I look forward to taking those classes -______-
Wish my professor was this clear helped a lot thanks
+SimplyGrouch Same
Stop being a weeb, things will become much clearer.
@@SlazeM7 HAHAHAHAHAHAHAA
Thank you for going the extra step and posting these videos. I really appreciate your style and approach, it is very refreshing.
Best of luck
Casey
I wish you were my actual teacher but still I am thankful to god to have access to this
This awesome. Teaches better than my actual professor. Thanks
### Calculus 1 Lesson 3.3: The First Derivative Test for Increasing and Decreasing Functions
Introduction
● [0:00] Professor Leonard introduces the topic: the first derivative test to determine intervals of increase and decrease of a function.
○ Mention of the application of this concept in curve sketching and understanding the shape of curves.
Review of Previous Concepts
● [0:55] Review of the meaning of the first derivative and its relationship with a function’s increase and decrease.
○ [1:10] If 𝒇'(x) > 0, the function 𝒇(x) is increasing.
○ [2:16] If 𝒇'(x) < 0, the function 𝒇(x) is decreasing.
○ [2:33] If 𝒇'(x) = 0, the function 𝒇(x) has a horizontal tangent, and a critical number is identified.
The First Derivative Test
● [3:05] Introduction to the first derivative test to find relative extrema (maximums and minimums).
● [3:26] Step-by-step explanation of the first derivative test:
○ 1.[4:21] Find the first derivative, 𝒇'(x).
○ 2.[4:40] Set the first derivative equal to zero, 𝒇'(x) = 0, to find critical numbers.
■ Besides the values where 𝒇'(x) = 0, the following points should also be included as “critical numbers,” where 𝒇'(x) does not exist:
◆ Points where the denominator is zero (possible vertical asymptotes).
► Occurs when the function is not defined (or tends to ±∞) as x approaches a certain value x = a.
An example is 𝒇(x) = 1/x. As x approaches 0, 𝒇(x) tends to ±∞, and the line x = 0 is a vertical
asymptote. Here, the function does not even reach a finite value at x = 0; it simply blows up to
infinity (or negative infinity).
◆ Sharp corners or cusps (abrupt changes in slope).
◆ Vertical tangents (infinite slope).
► Occurs when the function is (or can be) defined at that point, but its slope (derivative)
becomes infinite (or tends to ±∞). A typical example is 𝒇(x) = x^(1/3) at x = 0. The function
is indeed defined at x = 0, but its derivative 𝒇'(x) = 1 / (3x^(2/3)) tends to ±∞ as x approaches 0,
indicating a vertical tangent.
◆ Discontinuities (jump, infinite, removable).
○ 3. [4:58] Create a first derivative chart with critical numbers in order.
■ Draw a number line and mark the critical numbers.
■ Place 𝒇'(x) above the number line.
○ 4. [6:16] Test a point in each interval defined by the critical numbers.
■ Determine the sign of 𝒇'(x) in each interval.
◆ Substitute a value from each interval into 𝒇'(x) to determine if the slope is positive or negative.
■ The sign of 𝒇'(x) indicates whether the function is increasing or decreasing in each interval.
◆ Local Maximum:
► If 𝒇'(x) changes from positive to negative at a critical number, there is a local maximum.
◆ Local Minimum:
► If 𝒇'(x) changes from negative to positive at a critical number, there is a local minimum.
◆ Neither Maximum nor Minimum:
► If 𝒇'(x) does not change sign at a critical number, it is not classified as a maximum or minimum.
△ Sometimes referred to as a "horizontal inflection point" if:
▢ 𝒇'(x) = 0.
▢ The concavity changes (related to the second derivative).
Example 1: 𝒇(x) = x³ -3x +1
● [10:44] Application of the first derivative test to an example.
○ Find the derivative of the function.
○ Factor the derivative to find critical numbers.
○ Check for values that make the derivative undefined (denominators).
○ Create a first derivative chart with critical numbers.
○ Test a value from each interval in 𝒇'(x) to determine the sign.
○ Interpret the results to determine intervals of increase and decrease.
○ [14:13] Identify relative extrema (maximums and minimums).
○ Find the points of the relative extrema by substituting x-values into the original function.
Example 2: 𝒇(x) = 3x^(5/3) - 15x^(2/3)
● [17:12] Presentation of a second example to apply the first derivative test.
○ Find the first derivative of the function.
○ Factor the derivative to find critical numbers.
○ Check for values that make the derivative undefined (denominators).
○ Include values that make the denominator zero in the first derivative chart.
○ Create the first derivative chart with critical numbers, including values that make the denominator zero.
○ Test a value from each interval in 𝒇'(x) to determine the sign.
○ [23:05] Interpret the results to determine intervals of increase and decrease.
○ Identify relative extrema (maximums and minimums), considering values that make the denominator zero.
○ Determine whether the relative extrema are maximums or minimums.
Nobody talking abt Prof.'s humour👑
your videos are very straight forward . thank you for sharing the videos .
He's simply the best.
Super helpful for my Ap Calc class, thanks
thank you for uploading these videos they help a lot
Hor thing and lesson was good :)Thanks superprofessor
wish I found this professor earlier. My current one is just making calculus way more complicated than this.
1:15 I'll burn your paper and hand you back the staple... LOL
this man is a goddamn american hero
THANK YOU PROFESSOR!
ممتاز هالدكتور
جدا ومسلي
so great!
It's lucky for not having to take this test in the class :D
thank youuu
Professor Leonard drop the workout split
i took x^(2/3) as a factor and come up with the same answer on 18:33
I wish my teacher was also jacked maybe i would have focused more
hey, i wanted u to do deltan and epsilon form limit defination,
please
@ 9:23 Professor why are you flipping me off? lol
at 4:21 function must be continuous at critical points (where f' is zero or not defined). Your rule is true only for differentiable functions. You restrict the class of functions.
If a function is differentiable, the function MUST be continuous. Existing critical points means that the function has a derivative or is differentiable. Eventually, you are not true.
Starting around 17 minutes...I have a problem with him assigning a positive value to points with negative values under a radical. Seems like positive/imaginary for anything less than 0
This concerns me too. I agree with your point.
@@actuary32174 idk if I misunderstood you or not but if you mean the radical in the second question its odd radical so it's legal to put negatives there
@@moratasa9741ah that makes sense. Thanks
Try substituting -8. It's a cubed root not a square root, so the cubed root of -8 is -2. (Check that (-2)^3=-8). When you plug -8 into the numerator, you also get a negative answer. A negative numerator and a negative denominator give a positive final answer.
for the function f(x) = 3x^(5/3) - 5x^(2/3): We determined in your video that the slope is undefined at x=0. However the original function is continuous at x=0. What does this mean?
If you look at the graph of the function, you can see that the slope is becoming vertical as we approach a sharp turn at x=0 . The slope of a vertical line is undefined, so that is why the first derivative is undefined at x=0 even though the function is continuous there. (The sharp turn also makes the derivative not exist at that point, this function just happens to have both a sharp turn and a vertical slope at the same point)
But at 22:39 cube Root of x should not be equal to zero.
My goat
@21:42 if x = 0 doesn’t it mean that graph has a vertical tangent ?
yes and it can also possibly change signs (slope from decreasing to increasing and vise versa) there
How did you get the x^-1/3? How would 5^-1/3 * x = 5^2/3s?
Because x has an exponent of 1 therefore when factored it would be 1+-1/3=2/3 leaving it with 5^2/3
how did you pull out X ^-1/3 from X^2/3 and are left with an x?
Thank you! that really helped.
How can f(x) = 3x^(5/3) - 15x^(2/3) be a function when it has sharp turns at f(0?
Functions are allowed to have sharp turns. The absolute value function f(x)=|x| is another example of a function that has a sharp turn at x=0. For the function in the lecture, there is a sharp turn at x=0 but the function is still continuous there. However if you look at the graph you can see that the slope gets steeper and steeper as we approach x=0. The slope becomes vertical, and the slope of a vertical line is undefined. Further, the slope is approaching positive infinity as we approach x=0 from the left, and the slope approaches negative infinity as we approach from the right. So even though the function is continuous at x=0, we see that the first derivative is undefined there so the function is not differentiable at x=0. That doesn't stop us from finding the derivative at the other points where it is defined.
here where Algebra is crutia and will not play with u! it you miss your critical numbers your answer will be in a critical condition and you will find yourself trying to catch up some points at the hospitals problems :D
f(x)=4x+2/2x-3 (4x plus 2, over 2x minus 3). The (1st derivatives) f'(x)= -16/(2x-3)^2 (negative 16 over 2x minus 3 raise to the power of 2), and the f"(x)= 64/(2x-3)^3 When ask when is f'(x) is increasing or decreasing. How to compute when the numerator has no x value). In (second derivatives) f"(x), how to compute when f(x) is positive(conveks) or when it is negative (concave)? Can someone at least explain the logic behind it? Eventually youtube link if there is any.
May I have the ISBN code for the calc text you are using.
Stupid question but is this the explanation for section 3.3 James Stewart 8th edition "how derivatives affect the slope of the graph"
Yes
2 to 3
Are these videos for first year university calculus?
I think so, yes.
I studied pretty much this exact content when I studied "Differential Calculus" on my first year in Engineering Physics.
Then my final math course that year was "Multivariable Calculus", which is basically Calculus 3.
what edition is this Prof. Leonard? because i have edition 10th in my calculus and the chapters are different
bro why's that one kid always coughing in the back
My fiancee also hates math.
I would die for you
my math teacher should kiss your shoes
16:10 am i the only one heard the fly?
ينطي واهس يقرا 😂
Bro forget the calculus how can I have his physique 💀
He dabs at 24:37
lol @ how to find hors
+kinetic static go to the nearest hooker place
lolllll
bruh wear a full shirt, your biceps are too distracting(in a good way(like a compliment, but like, bad for me))