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anonymous
 4 years ago
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
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do first derivative test?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Goten77 ok for first derivative i got 15x^4  15x^2

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1k thats right now factor it

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1set 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
 4 years ago
Best ResponseYou've already chosen the best response.1now set up the critical points to test for when its positive and negative and you use the 1st derivative when plugging in theese valuesdw:1358134521331:dw u just use values in between the numbers

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1so 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
 4 years ago
Best ResponseYou've already chosen the best response.1part b is 2nd derivative test

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for part a.) is it (infinity,1) & (1,infinity)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok for part b.) i got x=0, and \[\pm \frac{ \sqrt{2} }{ 2 }\]

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1358135162509:dw hmm you number works to...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then i do the SDT? (Second Derivative Test)?

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1well there the same i meant XD

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1yes now you test those numbers between/around them

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, how do i do part c.)? o:

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1so um heres part bdw:1358135773604:dw

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1hmm the squareroot number is very small....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for b.) did u get \[(\frac{ \sqrt{2} }{ 2 }, \infty)\]

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1oh i keep on thinking about point of inflection <.<

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358128938034:dw o.O

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is that drawing seem right? :\

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1yes your right... its only concave up on that last interval

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so how do i do part c.) again sorry..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait so my part b.)was wrong?

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1so heres what the graph looks like dw:1358136607164:dw

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1aah i meant to say for part (c)

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1k everything is right now.... i just got part letters mixed up

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so how do ido part c.)

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1......u kno what i meant... that entre paragraph just now is part C

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1oh 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
 4 years ago
Best ResponseYou've already chosen the best response.0ah ok ty so for part c.) i got y=2x+2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah it says write the equation

Goten77
 4 years ago
Best ResponseYou've already chosen the best response.1k 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
 4 years ago
Best ResponseYou've already chosen the best response.0ohso my answer waswrong xx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@Goten77 canu help me with another 2 more pls? ^^
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