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