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anonymous
 5 years ago
please help to find cubic equation when f' (x) =0 at(1;3,5)
f'(x)<0 for x<1 and x>2
f'(x)>0 for 1<x<2
f(0) =0
f(2)=0 and f(4)=0
anonymous
 5 years ago
please help to find cubic equation when f' (x) =0 at(1;3,5) f'(x)<0 for x<1 and x>2 f'(x)>0 for 1<x<2 f(0) =0 f(2)=0 and f(4)=0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0use cardano's method given below

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok you know your derivative is negative then positive then negative, so it is a parabola facing down. the zeros are 1 and 2 so it looks like \[f'(x)=a(x+1)(x2)\] where a is some negative number. could even be 1 who knows.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0is the derivative the cubic? it has 3 zeros...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f'(x)=ax^2ax2a\] and therefore \[f(x) = \frac{ax^3}{3}\frac{ax^2}{2}2ax+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i assume this is looking for the cubic so the derivative is quadratic. with zeros at 1 and 2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0"please help to find cubic equation when f' (x) =0 at(1;3,5)" ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i am assuming you are looking for the cubic function whose derivative is zero at 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in other words \[f(x)=a_3x^3+a_2x^2+a_1x+a_0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[f(1)=3.5\] \[f(0)=0\] which of course means \[a_0=0\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0a quartic f(x) can have 3 zeros if its a touch and turn, coupled with a thru and thru

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0that would have 3 f' zeros then

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is a cubic, and they have given us the zeros. the original function is a cubic. the derivative is quadratic. the zeros of the original function are 0, 2 and 4, which means it is \[f(x)=a(x+2)(x4)x\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ho ho , that is a 4th degree poly. this one is sposed to be cubic!

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0how does a cubic have 3 zeros in the derivative? unless the original question is in error....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the derivative only has two zeros: 1 and 2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0"please help to find cubic equation when f' (x) =0 at(1;3,5" from the original statement ... I find this to either be an error, or the intended question...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is a parabola facing down, negative until 1, positive between 1 and 2, then negative

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0f' = 0 at 1,1,5 is my best interp of it.... right or wrong lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it is not error. what it means is that f'(1) = 0 and f(1)=3.5

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ooohhh....... that helps, thnx :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now let me see if i can finish this. before the day is over. lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok where were we before we were interrupted?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh yes, you have a cubic polynomial with zeros at 2,0, and 4 so it must look like \[f(x)=a(x+2)(x4)x\] where a is some negative number. negative because from the sign of the derivative we know that this polynomial is decreasing then increasing then decreasing. so we just need to find a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The question says that f'(x) is 0 at 1,3 and 5. But f'(1)=0 yields a=0!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i guess i cannot read. i read it as \[f'(1)=0\] at \[(1,3.5)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes but \[f^\prime(1)=0 \implies a=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually i get that too. damn

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the original quest. says find the equation of the cubic function when f'(x) = 0 at (1;3.5) then f'(x)<0 for x<1 and x>2 f'(x)>0 for 1<x<2 f(0) =0 f(2)=0 and f(4)=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in fact it get \[6a=8a\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@satellite73 that means a=0.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0think too many constraint to solve this one. yes, know it means a=0 that was what was bothering me.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if a = 0, just use a different letter lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lets go slow. do we agree that \[f(x)=a(x+2)(x4)x\] ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because f is cubic and the roots are given.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The cubic is \[f(x)=ax(x+2)(x4)\].

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so far so good. and we agree that a < 0 yes?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thx a mil for the awesome insight, any chance i would be able to graph this equation.....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because of the behavior of the derivative: negative then positive then negative means f is decreasing then increasing then decreasing. so a < 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0to show max an min values and intercepts

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and we also agree that \[f(x)=ax^32ax^28ax\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know what they want you to graph, but i think this problem is crap. could be wrong. they want you to graph a cubic that is decreasing until 1, then increasing until 2, then decreasing. and they want it to cross the x axis at 2, 0, and 4.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when f'(1) = 0 and f(3.5)=0 f'(x) =  ; (inf,1) U (2,inf) f'(x) = +; (1,2) <.......3.5..................1.....0............2............> + 0 + 0   0 + ????????

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0little backwards, but still i quandry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think 1,3.5 and the f'(x) being 0 date are wrong.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it says f'(1)=0 it changes direction at (1,3.5)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0f'(x)=0 could also indicate an inflection point... right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is entirely possible that this thing is over determined. you have not only the zeros of the cubic, but also the zeros of its derivative. i will try it later because i gotta go, but i think there is a contradiction here.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it is not an inflection point because the derivative changes sign there.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if the slope is the same on both sides of f' = 0, that indicates inflection i believe

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the signs dont change at 3.5

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0derivative changes sign. that is what it says.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is inflexion point noy determind by f''?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0in inflection point is made sure by f''; but an inflection point can still have a 0 slope in f'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0says f'(x)<0 for x<1 f'(x)>0 for 1<x<2 f"(x)< 0 for x>2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0derivative changes sign so f changes direction from decreasing to increasing. still i cannot come up with a. will try later.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0"when f'(x) = 0 at (1;3.5) then f'(x)<0 for x<1 and x>2 f'(x)>0 for 1<x<2"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.03.5 is an inflection point..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thx for all the help, egan helmie, south africa grade 12 2011provincial assesment task

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I see f(x) looking somthing like this if i interpret your info correctly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh no amistre, this is not a cubic polynomial. i interpret the question to be for a cubic polynomial. cubics do not look like this. also the second derivative of a cubic polynomial is a line, so only one inflection point, where the line crosses the x  axis (i.e. f''(x)=0). i am thinking this question was made up without sufficient thought. you cannot specify the zeros of a cubic polynomial, its value at some point, AND the zeros of the derivative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in fact i assert that there is no polynomial of degree 3 with zeros at 2, 0 and 4 for which f'(1)=0. let me see if i fall flat on my face with a (rather simple) proof. by the factor theorem we know that \[f(x)=ax(x+2)(x4)=ax^32ax^28ax\] thus \[f'(x)=3ax^24ax8a\] and \[f'(1)=3a+4a8a\] the only way for this to be 0 is for \[a=0\] which is a contradiction.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i trust your interpretations and its method; I dont trust that the given information is rigid enough to come to a solid conclusion tho ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is always possible i misinterpreted the question. but it is also possible some math teacher said "gee i will give them a problem to see if they can graph it" without thinking through whether such a thing exists, and that is the last second i am going to think about this.
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