Properties of Functions - Increasing vs Decreasing (Precalculus - College Algebra 9)

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How to determine intervals of Increasing and Decreasing from a graph.

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Professor Leonard, thank you for a well explained lecture on increasing and decreasing functions. I never really understand this topic in precalculus over the years, however your explanation is excellent from start to finish. Practice is the key to understanding the material.

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Thanks professor Leonard. A quick question, is PreCalculus the same as Advance Functions course? Thank you.

[@Umar-yi2ot]

So to sum it all up. Whenever the gradient > 0 the graph is increasing. And when the gradient < 0 the graph is decreasing.

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! I finally understand!!!!

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Thanks professor Leonard, any recommendations on a textbook for a self learner?

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For pre calculus

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Workout routine?

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complete

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Could you make some videos of linear transformations?.

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hows it going

[@Euphemia-em8fe]

day 2

[@alistaircornacchio5727]

Can someone clear up some confusion I have? I've seen this a few times already.. near the end of the video when he is doing that last graph.. he assumes it to be a vertical asymptote. How can we assume that? The output of the function could very well pass 0 and go into the negatives right? we wouldn't know. Is there some fundamental knowledge of what the arrows mean on a line that I am missing? pls help :)

[@anamikajha7210]

Dear Bro or Sis! u have a genuine question, however it needs some base. Keep reading. sir didn't assume it a vertical asymptote, it actually is. yes we truly have some basic knowledge for the arrow notations: this arrow shows that the graph is approaching the y axis, but it is not( don't get tensed, you'll get it) . Let me simplify what am tryna' expressing, try feeling it! what is a vertical asymptote? well, it's an *imaginary* line that cannot be crossed, but can be approached, mark it, can't be crossed but can be approached, just like u approach infinity. Now, the graph that we have is going to negative infinity vertically, but what will be the input or the x value for that? ( Now this is the point where u might have got a slap, open your eyes wide and read it with ur backbone as straight as possible) well, we can possibly observe that there is no y-intercept for x=0, so what does this reflect? again, WE CAN OBSERVE THAT THERE IS NO Y-INTERCEPT AS THE GRAPH ENDS WITH AN ARROW AS IT'S TAIL, SHOWING THAT "BYE!! MEET YOU AT INFINITY" , so what does this reflect? this reflects that for some x value the y value will be negative infinity, such that the x value is not negative ( READ ALOUD: the x value will not be negative because there is no y intercept, on a possible y value before infinity) so, it is your intuition that if, for example, x=1 and x=0.5 , then from the graph, we can see that for lesser input values, the output is approaching (learn this word by heart, thank me later) the y axis, so clearly at 0.5 that graph will be closer to the y axis. Now, at the same time , the y value for 0.5 will be closer to negative infinity, compared to the y value of 1, similarly 0.2 will be even closer to neg. inf. and 0.000000000000001 will again be closer to infinity than any of the above, similarly if x=0, will be negative infinity itself. Hence, we consider this a vertical asymptote because here the y value will be approaching negative infinity , which is impossible, and hence for this graph putting 0 is also approachable not reachable. hence it is a vertical asymptote. Sorry to make you mad. Read this word by word and you'll get it.

[@alistaircornacchio5727]

@@anamikajha7210 Hi thanks for the response, I appreciate it, however I'm not sure it answers my question. I think you are just explaining what an asymptote is, which I'm familiar with, but what I don't understand is how that is represented in his graph. Yes, I see that the arrow is pointing down to negative infinity, but there's no equation for the function so how can we assume that it never crosses x=0? I'm sure it's just a quick representation of a specific function, but it seems to me that we can't actually know (from the representation of the graph alone and not any further assuming) whether or not that line passes x=0 or not. Does my question make sense?

[@rosepierce9382]

@@alistaircornacchio5727 Yes, I was wondering about the same thing. I think the equation for the graph would actually make clear that it has a vertical asymptote, but it is not given here. Let me know if you have come across other satisfying explanations.

[@issamalsaid8501]

Hi. Prof...Why is the union signal placed in intervals of increasing and decreasing ?

[@Ironlionm4n]

I guess because they are sets of intervals

[@TwentyThree-23]

HELP!! At 1:48 wouldn’t it be X1 > X2 FOR DECREASING? Because if x1 is larger the At e,and x2 is smaller...

[@rosepierce9382]

As the input (x-value) increases, the output (y-value) decreases. That's what decreasing means. You go left to right checking whether the graph is falling (decreasing) or climbing (increasing). Two years late, but hope that clears it. How are you doing now? Did you master Precalculus? Passed the class/test/exam? I hope you're doing alright.

[@FirstLast-tj4nl]

Wouldn't the last graph be (-infinity, infinity) ? It never touches zero, so why include zero

[@alex-ie8tp]

(-infinity, infinity) would mean x goes towards -infinity and as you can see from that graph it never goes lower than 0, it just gets really really close to 0 but never reaching it (that's called an asymptote) which is why we have open parantheses (0, infinity)

[@bekastayib3310]

Where are your students? Keep up the good work

[@LalitKumar-os8si]

please help me to find the value of x (e^-x)-x=0

[@timetobesmart5140]

There is no such x x is (pn) such that e to the power of x=0

[@LalitKumar-os8si]

@@timetobesmart5140 thankyou for your reply but i know X = 0.56714329 approximately but i don't know how to find this

[@timetobesmart5140]

Lalit Kumar oh

[@dashaagafonova519]

Да да, фактических величин.

[@rikartc1728]

I've been doing some extra reading about increasing and decreasing functions, and it seems to be the case that there are functions that can be classified as "strictly increasing" and functions that can be classified as just "increasing". Basically the best way I can define the strictly increasing relationship of a function is that it is a one-to-one (injective) function, that for every input x it holds the following condition - f(xₙ) < f(xₙ₊₁). However, for just increasing functions, the following condition holds true for all inputs x - f(xₙ) ≤ f(xₙ₊₁), basically meaning that a constant can be classified as an increasing function. Am I completely off base here?