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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.

Mathematics
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1st derivative test?
do first derivative test?
@Goten77 ok for first derivative i got 15x^4 - 15x^2

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Other answers:

k thats right now factor it
|dw:1358134355078:dw|
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
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
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
part b is 2nd derivative test
|dw:1358134818974:dw|
for part a.) is it (-infinity,-1) & (1,infinity)
yes
ok for part b.) i got x=0, and \[\pm \frac{ \sqrt{2} }{ 2 }\]
|dw:1358135080380:dw|
|dw:1358135162509:dw| hmm you number works to...
then i do the SDT? (Second Derivative Test)?
well there the same i meant XD
yes now you test those numbers between/around them
ok, how do i do part c.)? o:
i told u part c
so um heres part b|dw:1358135773604:dw|
hmm the squareroot number is very small....
for b.) did u get \[(\frac{ \sqrt{2} }{ 2 }, \infty)\]
oh i keep on thinking about point of inflection <.<
|dw:1358128938034:dw| o.O
is that drawing seem right? :\
|dw:1358136164439:dw|
yes your right... its only concave up on that last interval
:"D
so how do i do part c.) again sorry..
wait so my part b.)was wrong?
so heres what the graph looks like |dw:1358136607164:dw|
aah i meant to say for part (c)
k everything is right now.... i just got part letters mixed up
so how do ido part c.)
......u kno what i meant... that entre paragraph just now is part C
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
ah ok ty so for part c.) i got y=-2x+2
no i gave u part 2
part c*
yeah it says write the equation
ill explain it again
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
ohso my answer waswrong x-x
okty :P
@Goten77 canu help me with another 2 more pls? ^^
sure

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