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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...
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infinity21
Canada6683 Posts
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)
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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 +
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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)
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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.
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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
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KK no need to make a mess for nothing missunderstanding happens.
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Better not be a calculus finals :D
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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//
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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
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EPT
![[image loading]](http://img78.imageshack.us/img78/4975/clazznu2.jpg)
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