lilfayfay
Let f be the function defined by f(x)=3x^5 - 5x^3 +2
(a) On what intervals is f increasing?
(b) On what intervals is the graph of f concave upward?
(c) Write the equation of each horizontal tangent line to the graph of f.
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Goten77
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1st derivative test?
lilfayfay
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do first derivative test?
lilfayfay
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@Goten77 ok for first derivative i got 15x^4 - 15x^2
Goten77
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k thats right now factor it
Goten77
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|dw:1358134355078:dw|
Goten77
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set each = to 0 and get critical points |dw:1358134445508:dw| but also find where x never exist.. but it always does so ignore that part
Goten77
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now set up the critical points to test for when its positive and negative
and you use the 1st derivative when plugging in theese values|dw:1358134521331:dw|
u just use values in between the numbers
Goten77
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so thats part a when its increasing/decreasing....
also part c is just the graphs of y = 0 , -1, +1 since those are the 3 horizontal tangents
Goten77
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part b is 2nd derivative test
Goten77
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|dw:1358134818974:dw|
lilfayfay
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for part a.) is it (-infinity,-1) & (1,infinity)
Goten77
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yes
lilfayfay
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ok for part b.) i got x=0, and \[\pm \frac{ \sqrt{2} }{ 2 }\]
Goten77
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|dw:1358135080380:dw|
Goten77
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|dw:1358135162509:dw| hmm you number works to...
lilfayfay
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then i do the SDT? (Second Derivative Test)?
Goten77
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well there the same i meant XD
Goten77
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yes now you test those numbers between/around them
lilfayfay
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ok, how do i do part c.)? o:
Goten77
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i told u part c
Goten77
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so um heres part b|dw:1358135773604:dw|
Goten77
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hmm the squareroot number is very small....
lilfayfay
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for b.) did u get \[(\frac{ \sqrt{2} }{ 2 }, \infty)\]
Goten77
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oh i keep on thinking about point of inflection <.<
lilfayfay
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|dw:1358128938034:dw|
o.O
lilfayfay
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is that drawing seem right? :\
Goten77
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|dw:1358136164439:dw|
Goten77
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yes your right... its only concave up on that last interval
lilfayfay
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:"D
lilfayfay
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so how do i do part c.) again sorry..
lilfayfay
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wait so my part b.)was wrong?
Goten77
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so heres what the graph looks like
|dw:1358136607164:dw|
Goten77
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aah i meant to say for part (c)
Goten77
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k everything is right now.... i just got part letters mixed up
lilfayfay
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so how do ido part c.)
Goten77
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......u kno what i meant... that entre paragraph just now is part C
Goten77
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oh i said it kinda wrong earlier i meant part Cis when f(x)= -1,0,1 meaning plug those 3 values into the original function and thats when the graph has a tangent lind so then you get y= 4 y= 2 y= 0 and thats when the graph has the horizontal tangents the x values of -1,0,1 came from the 1st derivative test since those are the critical points meaning where the derivative = 0
lilfayfay
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ah ok ty so for part c.) i got y=-2x+2
Goten77
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no i gave u part 2
Goten77
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part c*
lilfayfay
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yeah it says write the equation
Goten77
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ill explain it again
Goten77
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k so the 1st derivative gives us critical points meaning where the slope = 0 also meaning at those points on the original graph is where it has a horizontal tangent.....
so taht means the horizontal tangents are in the form of y = # and thats it
so our critical points were -1 0 and 1
plug those into the original equasion and get
y= 4
y= 2
y= 0
and thats it .. those are your 3 horizontal tangents
lilfayfay
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ohso my answer waswrong x-x
lilfayfay
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okty :P
lilfayfay
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@Goten77 canu help me with another 2 more pls? ^^
Goten77
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sure