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[H] Calculus workshop

Blogs > clazziquai
Post a Reply
clazziquai
Profile Blog Joined October 2007
6685 Posts
December 04 2008 07:43 GMT
#1
Okay this is like urgent, I try not to ask HW help threads but I have a finals tomorrow and I have a writeup due tomorrow.

There is a function f(x) = x^1/2 * (x - 1).
(a)What is the domain of f(x).
(b) What are the critical points of f(x). If there are no critical points write "None" .
(c) Find and label the local extrema (maxima and minima). If there are no local extrema
write "None" .
(d) What are the intervals of increase and decrease of f(x) = x^1/2 * (x - 1).
(e) Find the inflection points. If there are no inflection points write "N one"


a) The domain is [0, inf)

b) The critical point I found came out as c = 1/3 . Nothing else.

c) I am unsure of what to do. I basically do this:
f'(x) <------------(1/3)----------->
And find if any value less than 1/3 will be positive/negative and any value greater than 1/3 will be positive/negative. Then I can determine if there is a maxima/minima?

d) This is what confuses me. I have absolutely no clue on this.

e) None.

So basically c and d are the ones I have absolutely no clue on. Thank you so much! I will update as I keep trying to solve.
I need to get this over with to study for finals =/

#1 Sea.Really Fan / #1 Nesh Fan / Terran Forever~
blabber
Profile Blog Joined June 2007
United States4448 Posts
Last Edited: 2008-12-04 07:51:03
December 04 2008 07:50 GMT
#2
right for c...

you know what to do... so do it! Plug in something less than 1/3 for f'(x). See if it's positive or negative. Plug in something greater than 1/3 into f'(x). See if it's positive or negative.

remember the derivative is the slope of the original function!
if you see f'(x) +++ 1/3 ---- then it's a maximum (if you visualize it, a graph that has positive slope then becomes negative slope has a maximum!). then it's the opposite way for minimum

for d, from what I can tell they're just asking you where the graph increases and decreases, which is basically what you did in c...
blabberrrrr
infinity21 *
Profile Blog Joined October 2006
Canada6683 Posts
December 04 2008 07:52 GMT
#3
You can either approach your critical point from both sides and determine it that way or you can take the 2nd derivative and see the values of a small interval around 1/3 for c

d) Just draw it out. If 1/3 is the local max, then anything to the left of 1/3 is increasing, anything to its right is decreasing, etc. Alternatively, you can look at the values of the derivatives in the intervals created by the end points and each critical point
So look at the intervals [0, 1/3) and (1/3, inf)
Official Entusman #21
Malongo
Profile Blog Joined November 2005
Chile3472 Posts
Last Edited: 2008-12-04 09:00:08
December 04 2008 08:12 GMT
#4
There is a function f(x) = x^1/2 * (x - 1).
(a)What is the domain of f(x). ==> where exists f(x)
==> where exists f(x) => x>=0

(b) What are the critical points of f(x). If there are no critical points write "None" .
==> find f'(x)= 1/2*x^(-1/2)*(x-1) + x^(1/2) critic points where this is =0 or this doesnt exist
x=0 and x= 1/3 those BOTH


(c) Find and label the local extrema (maxima and minima). If there are no local extrema
write "None" .
posible local extrema = critic points
calculate f''(x)= 3/4*x^(-1/2) + 1/4*x^(-3/2) then put critic points here f''(1/3)>0 posible MIN (if <posible MAX)
how do you know if it is MIN compare with the other critic points hence you have 1/3 and 0
f(1/3)< 0 f(0)=0 1/3 is Min. theres no MAX (if multiple c p then pick all where f''>0 and compare on f, do the same to those with f''<0 to find max)


(d) What are the intervals of increase and decrease of f(x) = x^1/2 * (x - 1).
increase f'(x)>0 , decrease f'(x)<0 sol dec[0,1/3] inc[1/3,inf(


(e) Find the inflection points. If there are no inflection points write "N one"
inflection points f''(x)=0 sol inflection points NONE because f''(x) is never 0

bonus track+ Show Spoiler +

[image loading]

[image loading]
Help me! im still improving my English. An eye for an eye makes the whole world blind. M. G.
freelander
Profile Blog Joined December 2004
Hungary4707 Posts
Last Edited: 2008-12-04 08:38:47
December 04 2008 08:18 GMT
#5
On December 04 2008 16:50 blabber wrote:
right for c...

you know what to do... so do it! Plug in something less than 1/3 for f'(x). See if it's positive or negative. Plug in something greater than 1/3 into f'(x). See if it's positive or negative.

remember the derivative is the slope of the original function!
if you see f'(x) +++ 1/3 ---- then it's a maximum (if you visualize it, a graph that has positive slope then becomes negative slope has a maximum!). then it's the opposite way for minimum

for d, from what I can tell they're just asking you where the graph increases and decreases, which is basically what you did in c...


well.. for finding out local extremas there is a much better method.
the derivative is the slope of the function.. right.

and the second derivative is the slope of the first derivative ==> if it equals 0 somewhere, it means the first derivative doesn't increase or decrease in that point, because it changes from positive to negative or vice versa ==> it means the original function has a local minimum or maximum there, because if there is a local extrema, the function must increase/decrease before it and decrease/increase after it (first: max second: min)

you won't know after this if it is max or min though.


for d)
the derivative is the slope of the function..

if it is positive, the function is increasing,
if it is negative the function is decreasing.

e)
where f"(x)==0
the reason is what I wrote in c)
And all is illuminated.
Wonders
Profile Blog Joined September 2006
Australia753 Posts
December 04 2008 08:23 GMT
#6
For d: If f'[x] is positive for a particular x then f is increasing at x (Intuitively, if the rate of change of the function is positive, then the function is increasing). If it's negative then f is decreasing. So compute f' and find where it's positive and where it's negative.
blabber
Profile Blog Joined June 2007
United States4448 Posts
December 04 2008 08:47 GMT
#7
On December 04 2008 17:18 freelander wrote:
Show nested quote +
On December 04 2008 16:50 blabber wrote:
right for c...

you know what to do... so do it! Plug in something less than 1/3 for f'(x). See if it's positive or negative. Plug in something greater than 1/3 into f'(x). See if it's positive or negative.

remember the derivative is the slope of the original function!
if you see f'(x) +++ 1/3 ---- then it's a maximum (if you visualize it, a graph that has positive slope then becomes negative slope has a maximum!). then it's the opposite way for minimum

for d, from what I can tell they're just asking you where the graph increases and decreases, which is basically what you did in c...


well.. for finding out local extremas there is a much better method.
the derivative is the slope of the function.. right.

and the second derivative is the slope of the first derivative ==> if it equals 0 somewhere, it means the first derivative doesn't increase or decrease in that point, because it changes from positive to negative or vice versa ==> it means the original function has a local minimum or maximum there, because if there is a local extrema, the function must increase/decrease before it and decrease/increase after it (first: max second: min)

you won't know after this if it is max or min though.


So basically you're just saying what I'm saying... lol... I normally like to keep any talk of a second derivative out of finding that stuff because it gets somewhat confusing
blabberrrrr
freelander
Profile Blog Joined December 2004
Hungary4707 Posts
Last Edited: 2008-12-04 09:37:12
December 04 2008 09:12 GMT
#8
this post was wrong
And all is illuminated.
Malongo
Profile Blog Joined November 2005
Chile3472 Posts
Last Edited: 2008-12-04 09:39:50
December 04 2008 09:21 GMT
#9
KK no need to make a mess for nothing missunderstanding happens.
Help me! im still improving my English. An eye for an eye makes the whole world blind. M. G.
Nytefish
Profile Blog Joined December 2007
United Kingdom4282 Posts
December 04 2008 09:27 GMT
#10
Better not be a calculus finals :D
No I'm never serious.
clazziquai
Profile Blog Joined October 2007
6685 Posts
December 04 2008 09:28 GMT
#11
On December 04 2008 18:27 Nytefish wrote:
Better not be a calculus finals :D


haha nope, something different. i just hate having other hw to do when i have finals the next day haha//
#1 Sea.Really Fan / #1 Nesh Fan / Terran Forever~
freelander
Profile Blog Joined December 2004
Hungary4707 Posts
December 04 2008 09:36 GMT
#12
On December 04 2008 18:21 malongo wrote:
Show nested quote +
On December 04 2008 18:12 freelander wrote:
so you say there is no inflection point but there is one on your graphics..
also I think the domain is the whole IR


sure maybe the domain is IR... lets see if clazz can understand your post
and inflection... well maybe you can point the inflection point for me? i know tl homework threads are bad but..



sorry I edited out the domain thing after i thought about it second time.. ok the inflection point thing i was wrong
And all is illuminated.
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