Differentiation : Increasing & Decreasing Functions : ExamSolutions
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- Опубліковано 17 жов 2010
- Differentiation of functions and finding whether they are increasing or decreasing.
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Why is dy/dx greater than 0 when the equation is increasing
A positive gradient (going upwards) means a function is increasing and since dy/dx represents gradient then this will be positive, >0
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No. Linear equations have the form f(x) = mx + c.
It is increasing when 2x-4>0 so x>2 and decreasing when 2x-4
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Thanks for watching.
What would you do if you have to use the quadratic formula?
How would you determine whether a function is increasing or decreasing for all real values of x?
@Examsolutions how do i find the set of values of the equation since i do dy/dx of (x-2)^2 to get 2x-4 and there is only one value of x. Please help
Are the critical values just 'x' coordinates of the stationary points? Thanks for your videos, they're amazing!
In this case yes but there may not be stationary points in some cases so you will still need to be aware that for increasing functions dy/dx>0 and for decreasing functions, dy/dx
ExamSolutions Oh ok I understand. Thanks for all your help, it has been amazing!
@allstarjai1010 all the values between
Get more on Increasing and Decreasing functions:
*From Graphs( ua-cam.com/video/9G9EManhGbw/v-deo.html )
*Calculus Method ( ua-cam.com/video/XSKozWXwqXc/v-deo.html )
this is really helpful thank you
Thanks for your comment, best wishes
Get more on Increasing and Decreasing functions:
*From Graphs( ua-cam.com/video/9G9EManhGbw/v-deo.html )
*Calculus Method ( ua-cam.com/video/XSKozWXwqXc/v-deo.html )
god bless you
THANKSSS
Thank you for watching.
Get more on Increasing and Decreasing functions:
*From Graphs( ua-cam.com/video/9G9EManhGbw/v-deo.html )
*Calculus Method ( ua-cam.com/video/XSKozWXwqXc/v-deo.html )
in the exam can i just use my calculator to find what the graph looks like exactly and then just use that to figure out the answer? and by the way, your videos are so helpful i really appreciate the time and effort you put into making them.
mazin mohammed c1 doesn't allow calculators
Georgia Gill I'm doing C12
Is it one of those calculators where you can generate a graph? If so then I don't think that exam boards even allow those calculators unless you have an exam board which does.
it wont generate a graph it's simply going to give me the coordinates of the points that lie on that specific curve, from which I can figure out the shape of the graph.
And how can you do that ? Because i am giving C12 Exam too.
Where you've drawn the f(x) graph, how can you tell that its a u shape rather than n shape?
if its -x^2 then its n shape, if its positive x^2 then its u-shape
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and as always, awesome tutorial
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what if the equation cannot be factorised?
Rohit Attri exactly my point
Use quadratic formula
ua-cam.com/video/CA9eNbFW5SE/v-deo.html
can't understand in (3x-2)(x+4)>0, from where appeared (-2) and +4...
does't looks clear for me.
by factorization
How would you PROVE something is increasing/decreasing for all values of x? I’ve seen like 3 methods:
-find dy/dx, and complete the square and then a positive constant means dy/dx>0 and is increasing, and vice versus for decreasing
-find d2y/dx^2 but I don’t really understand how to do it this way,
-or find dy/dx, then set equal to 0 to find stationary point(s), then find gradient slightly to the right and left of them, and hence if all positive or negative, increasing/decreasing for all values of x.
-online I’ve seen some people find stationary points by factorising, then obviously showing both solutions are positive by finding the gradient to the left and right of the stationary points, and if increasing its all positive, and decreasing if all negative
What method is right or are they all valid? My textbook uses the first method but my teacher has said to use he third, and a friend told me to use the second, and I’ve seen some examples online of the fourth
@Examsolutions would my graph be linear is my equation is f(x) = (x-2)^2
Sorry, but how would this be solved
dy/dx = 2(18x-1)(3x-1)^4
Find the set of values for x for which dy/dx < 0
Thank you in advance
Since (3x-1)^4 is positive for all values of x then for dy/dx
Thank you very much for replying, I got the solution x
Are you sure the question is just lest than zero. It seems from what you are saying it is less than or equal to zero in which case 3x-1=0 leading to x=1/3. If this were the case x is less than or equal to 1/18
ExamSolutions Yes, I'm really sorry, it is less than or equal to zero. Thank you very much for the help
No problem. Yes it would have a gradient m=0 when dy/dx =0 but surely that is what is being asked in the question. They are asking for values of x for which dy/dx is less than or equal to zero.
@MEGIDIOT not a gud idea but ill do it aswell lool :P
do you like bladee?
did no one notice that he solved the quadratic wrong...
Sorry but he didn't