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anonymous
 one year ago
Let f(x)=3x^3−36x+3.
On which interval is f increasing? Decreasing?
Find the points at which f achieves a local min. and local max.
Find intervals on which f is concave up and concave down.
Find all points of inflection.
anonymous
 one year ago
Let f(x)=3x^3−36x+3. On which interval is f increasing? Decreasing? Find the points at which f achieves a local min. and local max. Find intervals on which f is concave up and concave down. Find all points of inflection.

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Jack1
 one year ago
Best ResponseYou've already chosen the best response.1f(x)=3x^3−36x+3 Find the points at which f achieves a local min. and local max. so solve for f'(x) = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[f'(x) = 9x^2 36\] Setting that equal to 0 I get x=2 and x=2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How do I know which is a max. and which is a min. ? Do I plug it into the original function?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3to find out if f is increasing or decreasing you plug in x=2 and x =2 into the first derivative of the equation

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3increasing and decreasing comes from the first derivative concave up and concave down comes from the second derivative.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3for increasing and decreasing if your result after plugging in x = 2 or 2 is negative, we are decreasing. If your result after plugging in x =2 or 2 is positive, we are increasing.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3local min and local max... well we need to grab the absolute max (highest point in the graph) and the absolute min (lowest point the graph) before dealing with that. local min is decreasing over a certain area in the graph and local max is increasing over a certain area in the graph

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3inflection point (omg hope I still remember this) is the point where the graph is switching from negative to positive or positive to negative.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434785484976:dw this is just an example...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3x_X! ok ok... wait ... x = 2 and x =2 are critical points... we have to create a chart.... dw:1434785632760:dw pick a point between those intervals.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3if it's negative it's decreasing and if it's positive it's increasing. pick a point between those intervals and plug it into the first derivative.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3do the interval where f is increasing or decreasing(first derivative) and the interval for concavity first (2nd derivative) first... then once you obtain those results, you can do the local min, local max, and points of inflection

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3you can only answer the questions for local min/max and inflection after you do the first question and the third question ...otherwise, you'll be stuck

princeharryyy
 one year ago
Best ResponseYou've already chosen the best response.0still need help ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I plug in x=2 or x=2 into the first derivative I will only get 0 again, because I set f'(x)=0 in order to get the critical numbers 2 and 2.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3you have to pick a point besides 2 and 2 in each interval

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434792689289:dw your critical points are x = 2, and 2 remember this chart? for each interval (oo,2), (2,2), and (2,oo) you have to pick a point.. like let x = 3 and plug it into the derivative equation similarly x = 1 and x =3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Right, so then if it becomes positive then it is increasing on that interval, and if negative, it is decreasing?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[f'(x) = 9x^2 36 \] first derivative plug in x = 3 for the interval (oo,2) x = 1 for (2,2) x=3 for (2,oo) yes @Tracy96

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so f'(3) = 45 (increasing) f'(1) = 27 (decreasing) f'(3) = 45 (increasing) Is that right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The calculator's giving me 27, is that wrong?

Jack1
 one year ago
Best ResponseYou've already chosen the best response.1Let \(f(x)=3x^3−36x+3\) Find the points at which f achieves a local min. and local max. find the derivative \( \large f′(x)=9x^2−36\) derivative = gradient... so when gradient = 0, its a turning poing \( \large f′(0)=9x^2−36\) \( \large f′(0)=2~ or~ 2\) so what are the y values at these points? \(f(x)=3x^3−36x+3\) \(f(2)=3x^3−36x+3\) \(f(2)=45\) = local minima \(f(2)=51\) = local maxima

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3sorry.. yes that's right.. 27 for f(1) I hit the wrong number by accident

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3dw:1434793225103:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3now we need the second derivative for concavity

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3\[\large f′(x)=9x^2−36\] take the derivative of this equation again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Jack1 Ahh I see! That makes sense!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@UsukiDoll Great! So the 2nd derivative is 18x right?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3yes... so wee need to find the critical point f''(x) = 18x set the equation to 0 and solve for x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3yeah... so there is only one critical point this time x =0 so we only find intervals for (oo,0) and (0,oo) pick a point.. now the only problem is I tend to have the concave up or concave down vocabulary mixed up :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Haha no worries... so can we choose x=1 and x=1 ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3but again depends on the number oh here it is if f''(x) > 0 concave up f''(x) <0 concave down.. yes we can choose those values as long as it's in the interval (oo,0) for x = 1 and (0,oo) for x =1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And I have to plug them in the 2nd derivative am I right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0f''(1) = 18 and f''(1) = 18

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3so for the interval (oo,0) we have concave down because f''(x) <0 concave down.. and for the interval (0,oo) we have concave up because f''(x) > 0 is concave up

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh okie dokie! That makes a lot of sense now! :D ahaha

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3now I sort of forgot how the drawing is.. well I know it .. but I keep switching them unintentionally

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By this point have we found the points of inflection, or is there more to calculate?

Jack1
 one year ago
Best ResponseYou've already chosen the best response.1Find intervals on which f is concave up and concave down. find the points either side of the local minima/maxima to work out if they're higher or lower of either side is lower, if the turning point is the highest point (highest y value), therefore it's concave up if the turning point is the lowest point (lowest y value), its concave down f(x)=3x^3−36x+3. 1st turning point = (2, 51) so a point either side of 2 is 1.9 and 2.1 y value at x = 1.9 \(\large f(x)=3x^3−36x+3\) \(\large f(1.9)=3x^3−36x+3\) \(\large f(1.9)=50.823\) y value at x = 2.1 \(\large f(x)=3x^3−36x+3\) \(\large f(2.1)=3x^3−36x+3\) \(\large f(2.1)=50.817\) each point either side is slightly lower than 51, so it's concave down do the same for each side of the other turning point (2,45) to work out what it is

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I don't think so ... but I think dw:1434793917549:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3what the heck? I think he's doing it a different way. The way I learned it is to create a chart and plug in values in the intervals after solving for 0 and finding the critical points.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm I understand :) @Jack1 That makes sense!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3a. find first and second derivatives b. find critical points c. do sign chart  plug in whatever value you pick in the first and second derivative.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0My prof is teaching me the way with the chart which confuses me at times, but the method @Jack1 is using was the 1st method I learned back in high school

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3why is the chart confusing? I'm curious... that's how my professor taught me and it was straight forward.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3actually I went into a debate with my Mathematical Biology professor last semester..because I used the sign chart and first derivative to find increasing and decreasing but he was all like derrr grab the second derivative and look at the sign. what is that? >:/

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I know there is more than one method to solve math problems as long as it doesn't break math rules... but creating math shortcuts like what my professor did (last semester) threw me off. But, I did used it in the midterms.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hahaha idk, I guess the 1st way you learn things always sits in the back of your mind better :D lol

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3hahahaa yeah... I still don't agree with his method XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I agree, it's quite hard to agree with profs sometimes :P haha

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I'm glad I don't have to deal with him anymore!!!!!!!!!!! :D not too many people are in his classes for the upcoming Fall semester. . . I can't wait until it gets canceled which most likely is! :D He was such a jerk. I don't care if I passed with an above average grade

Jack1
 one year ago
Best ResponseYou've already chosen the best response.1Find all points of inflection. Let \(\large f(x)=3x^3−36x+3\) find 2nd derivative \(\large f(x)=3x^3−36x+3\) \(\large f'(x)=9x^2−36\) \(\large f''(x)=18x\) when f''(x) = 0, it's a point of inflection \(\large f''(x) = 18x\) \(\large f''(0) = 18x\) \(\large f''(0) = 0\) ==> so x = 0 is a point of inflection what is the y value here? \(\large f(x)=3x^3−36x+3\) \(\large f(0)=3x^3−36x+3\) \(\large f(0)=3\) so point of inflection is at ( 0, 3)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hahahaha right on!! XD above average grade!! I hope the same happens with me! :D

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3whoa.. nice one @Jack1

Jack1
 one year ago
Best ResponseYou've already chosen the best response.1On which interval is f increasing? Decreasing? psshh... graph it or freehand draw it after u have worked out the other main points... if y value is increasing from left to right )if it looks uphill) then f is increasing

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3have you seen the questions from this one guy in the Qualified Helpers section? It's crazy! :/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Jack1 Ok so whatever x value you find when you set 2nd derivative = 0 is an inflection point?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3we have the interval increasing or decreasing already.. via the chart.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Jack1 haha yeah don't worry about the graphing... if I'm good at one thing , its graphing xD lol

Jack1
 one year ago
Best ResponseYou've already chosen the best response.1yep @Tracy96 ... 2nd derivative shows the rate of change of the gradient... so if it's 0, then at that point it's transitioning from a decreasing gradient to an increasing gradient, or vice versa

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahhh alrighty!! @Jack1 very helpful tips thanks so much!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And @UsukiDoll thank to you too!! I got everything you said! Including the 'chart' method! :D haha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And thanks for the medal! :D @UsukiDoll

Jack1
 one year ago
Best ResponseYou've already chosen the best response.1@UsukiDoll 's way of doing charts and finding the critical points will probably be the quicker way to do it overall... I just find it more logical in my mind to do the other way

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3xD thanks... everyone got a medal on here

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, logic is always a saviour when it comes to exams >.< lol thanks again!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@UsukiDoll mind me asking which school you go to?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah lol same thing here in Canada... Which year are you? I'm a 1st year xD

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I'm known as a super senior.. still going to school despite reaching over 120 credits

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well its gonna be my 2nd year this fall :D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh wow!!! 120 credits!!! Thats a lot!!!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I actually have 128...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3but I went to community college for 6 years because I was horrendous at Math, my community college screwed me over for having a requirement that's impossible to achieve...plus I was still iffy about what I wanted to do after college... but now I know and that's to teach... hopefully if I CAN GET OVER THE STUPID ANALYTICAL LOGIC MATH STUFF I can try Graduate school so I can teach adults.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I need a MA in Math to teach in a college campus ...otherwise I would be stuck with high schoolers.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hahahaha teaching high schoolers is a serious challenge XD lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Specially when it comes to Math lol

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3oh yeah my state is the worst.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And wow! 6 years of college! That should take a lot hard work!!

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.3I think part of it was do to me being in a different Major which messed up my gpa so I stopped enrolling in Business courses and focused on Math
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