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.

- anonymous

- katieb

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- Goten77

1st derivative test?

- anonymous

do first derivative test?

- anonymous

@Goten77 ok for first derivative i got 15x^4 - 15x^2

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## More answers

- Goten77

k thats right now factor it

- Goten77

|dw:1358134355078:dw|

- Goten77

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

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

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

part b is 2nd derivative test

- Goten77

|dw:1358134818974:dw|

- anonymous

for part a.) is it (-infinity,-1) & (1,infinity)

- Goten77

yes

- anonymous

ok for part b.) i got x=0, and \[\pm \frac{ \sqrt{2} }{ 2 }\]

- Goten77

|dw:1358135080380:dw|

- Goten77

|dw:1358135162509:dw| hmm you number works to...

- anonymous

then i do the SDT? (Second Derivative Test)?

- Goten77

well there the same i meant XD

- Goten77

yes now you test those numbers between/around them

- anonymous

ok, how do i do part c.)? o:

- Goten77

i told u part c

- Goten77

so um heres part b|dw:1358135773604:dw|

- Goten77

hmm the squareroot number is very small....

- anonymous

for b.) did u get \[(\frac{ \sqrt{2} }{ 2 }, \infty)\]

- Goten77

oh i keep on thinking about point of inflection <.<

- anonymous

|dw:1358128938034:dw|
o.O

- anonymous

is that drawing seem right? :\

- Goten77

|dw:1358136164439:dw|

- Goten77

yes your right... its only concave up on that last interval

- anonymous

:"D

- anonymous

so how do i do part c.) again sorry..

- anonymous

wait so my part b.)was wrong?

- Goten77

so heres what the graph looks like
|dw:1358136607164:dw|

- Goten77

aah i meant to say for part (c)

- Goten77

k everything is right now.... i just got part letters mixed up

- anonymous

so how do ido part c.)

- Goten77

......u kno what i meant... that entre paragraph just now is part C

- Goten77

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

- anonymous

ah ok ty so for part c.) i got y=-2x+2

- Goten77

no i gave u part 2

- Goten77

part c*

- anonymous

yeah it says write the equation

- Goten77

ill explain it again

- Goten77

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

- anonymous

ohso my answer waswrong x-x

- anonymous

okty :P

- anonymous

@Goten77 canu help me with another 2 more pls? ^^

- Goten77

sure

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