anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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Goten77
  • Goten77
1st derivative test?
anonymous
  • anonymous
do first derivative test?
anonymous
  • anonymous
@Goten77 ok for first derivative i got 15x^4 - 15x^2

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

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