anonymous
  • anonymous
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.
Mathematics
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Jack1
  • Jack1
f(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
  • anonymous
\[f'(x) = 9x^2 -36\] Setting that equal to 0 I get x=2 and x=-2
anonymous
  • anonymous
How do I know which is a max. and which is a min. ? Do I plug it into the original function?

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UsukiDoll
  • UsukiDoll
to find out if f is increasing or decreasing you plug in x=2 and x =-2 into the first derivative of the equation
UsukiDoll
  • UsukiDoll
increasing and decreasing comes from the first derivative concave up and concave down comes from the second derivative.
UsukiDoll
  • UsukiDoll
for 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
  • UsukiDoll
local 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
  • UsukiDoll
inflection point (omg hope I still remember this) is the point where the graph is switching from negative to positive or positive to negative.
UsukiDoll
  • UsukiDoll
|dw:1434785484976:dw| this is just an example...
UsukiDoll
  • UsukiDoll
x_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
  • UsukiDoll
if 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
  • UsukiDoll
do 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
  • UsukiDoll
you 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
  • princeharryyy
still need help ?
anonymous
  • anonymous
If 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.
anonymous
  • anonymous
@UsukiDoll
Jack1
  • Jack1
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Jack1
  • Jack1
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Jack1
  • Jack1
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UsukiDoll
  • UsukiDoll
you have to pick a point besides -2 and 2 in each interval
UsukiDoll
  • UsukiDoll
|dw: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
  • anonymous
Right, so then if it becomes positive then it is increasing on that interval, and if negative, it is decreasing?
UsukiDoll
  • UsukiDoll
\[f'(x) = 9x^2 -36 \] first derivative plug in x = -3 for the interval (-oo,-2) x = 1 for (-2,2) x=3 for (2,oo) yes @Tracy96
anonymous
  • anonymous
Ok so f'(-3) = 45 (increasing) f'(1) = -27 (decreasing) f'(3) = 45 (increasing) Is that right?
anonymous
  • anonymous
The calculator's giving me -27, is that wrong?
Jack1
  • Jack1
Let \(f(x)=3x^3−36x+3\) -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
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UsukiDoll
  • UsukiDoll
sorry.. yes that's right.. -27 for f(1) I hit the wrong number by accident
UsukiDoll
  • UsukiDoll
|dw:1434793225103:dw|
UsukiDoll
  • UsukiDoll
now we need the second derivative for concavity
UsukiDoll
  • UsukiDoll
\[\large f′(x)=9x^2−36\] take the derivative of this equation again
anonymous
  • anonymous
@Jack1 Ahh I see! That makes sense!
anonymous
  • anonymous
@UsukiDoll Great! So the 2nd derivative is 18x right?
UsukiDoll
  • UsukiDoll
yes... so wee need to find the critical point f''(x) = 18x set the equation to 0 and solve for x
anonymous
  • anonymous
x=0 ?
UsukiDoll
  • UsukiDoll
yeah... 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
  • anonymous
Haha no worries... so can we choose x=-1 and x=1 ?
UsukiDoll
  • UsukiDoll
but 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
  • anonymous
And I have to plug them in the 2nd derivative am I right?
UsukiDoll
  • UsukiDoll
yeah
anonymous
  • anonymous
f''(-1) = -18 and f''(1) = 18
UsukiDoll
  • UsukiDoll
so 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
  • anonymous
Ohhh okie dokie! That makes a lot of sense now! :D ahaha
UsukiDoll
  • UsukiDoll
now I sort of forgot how the drawing is.. well I know it .. but I keep switching them unintentionally
anonymous
  • anonymous
By this point have we found the points of inflection, or is there more to calculate?
Jack1
  • Jack1
-Find 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
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UsukiDoll
  • UsukiDoll
I don't think so ... but I think |dw:1434793917549:dw|
UsukiDoll
  • UsukiDoll
what 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
  • anonymous
Hmmm I understand :) @Jack1 That makes sense!
UsukiDoll
  • UsukiDoll
a. 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
  • anonymous
My 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
  • UsukiDoll
why is the chart confusing? I'm curious... that's how my professor taught me and it was straight forward.
UsukiDoll
  • UsukiDoll
actually 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
  • UsukiDoll
I 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
  • anonymous
Hahaha idk, I guess the 1st way you learn things always sits in the back of your mind better :D lol
UsukiDoll
  • UsukiDoll
hahahaa yeah... I still don't agree with his method XD
anonymous
  • anonymous
LOL XD
anonymous
  • anonymous
I agree, it's quite hard to agree with profs sometimes :P haha
UsukiDoll
  • UsukiDoll
I'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
  • Jack1
-Find 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
  • anonymous
Hahahaha right on!! XD above average grade!! I hope the same happens with me! :D
UsukiDoll
  • UsukiDoll
whoa.. nice one @Jack1
Jack1
  • Jack1
-On 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
  • UsukiDoll
have you seen the questions from this one guy in the Qualified Helpers section? It's crazy! :/
anonymous
  • anonymous
@Jack1 Ok so whatever x value you find when you set 2nd derivative = 0 is an inflection point?
UsukiDoll
  • UsukiDoll
we have the interval increasing or decreasing already.. via the chart.
anonymous
  • anonymous
@Jack1 haha yeah don't worry about the graphing... if I'm good at one thing , its graphing xD lol
Jack1
  • Jack1
yep @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
  • anonymous
Ahhh alrighty!! @Jack1 very helpful tips thanks so much!!
anonymous
  • anonymous
And @UsukiDoll thank to you too!! I got everything you said! Including the 'chart' method! :D haha
anonymous
  • anonymous
And thanks for the medal! :D @UsukiDoll
Jack1
  • Jack1
@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
  • UsukiDoll
xD thanks... everyone got a medal on here
anonymous
  • anonymous
Yeah, logic is always a saviour when it comes to exams >.< lol thanks again!!
Jack1
  • Jack1
np =)
anonymous
  • anonymous
:D
anonymous
  • anonymous
@UsukiDoll mind me asking which school you go to?
UsukiDoll
  • UsukiDoll
university
anonymous
  • anonymous
yeah lol same thing here in Canada... Which year are you? I'm a 1st year xD
UsukiDoll
  • UsukiDoll
I'm known as a super senior.. still going to school despite reaching over 120 credits
anonymous
  • anonymous
Well its gonna be my 2nd year this fall :D
anonymous
  • anonymous
Oh wow!!! 120 credits!!! Thats a lot!!!
UsukiDoll
  • UsukiDoll
I actually have 128...
anonymous
  • anonymous
Lucky you!!! :O
UsukiDoll
  • UsukiDoll
but 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
  • UsukiDoll
I need a MA in Math to teach in a college campus ...otherwise I would be stuck with high schoolers.
anonymous
  • anonymous
Hahahaha teaching high schoolers is a serious challenge XD lol
anonymous
  • anonymous
Specially when it comes to Math lol
UsukiDoll
  • UsukiDoll
oh yeah my state is the worst.
anonymous
  • anonymous
And wow! 6 years of college! That should take a lot hard work!!
anonymous
  • anonymous
lot of*
UsukiDoll
  • UsukiDoll
I 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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