anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
use cardano's method given below
anonymous
  • anonymous
ok 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 \[f'(x)=a(x+1)(x-2)\] where a is some negative number. could even be -1 who knows.
amistre64
  • amistre64
is the derivative the cubic? it has 3 zeros...

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amistre64
  • amistre64
f' = 0 at -1,-3,5 ?
anonymous
  • anonymous
\[f'(x)=ax^2-ax-2a\] and therefore \[f(x) = \frac{ax^3}{3}-\frac{ax^2}{2}-2ax+c\]
anonymous
  • anonymous
good morning.
amistre64
  • amistre64
howdy :)
anonymous
  • anonymous
i assume this is looking for the cubic so the derivative is quadratic. with zeros at -1 and 2
amistre64
  • amistre64
"please help to find cubic equation when f' (x) =0 at(-1;-3,5)" ??
anonymous
  • anonymous
i am assuming you are looking for the cubic function whose derivative is zero at -1
anonymous
  • anonymous
in other words \[f(x)=a_3x^3+a_2x^2+a_1x+a_0\]
anonymous
  • anonymous
\[f(-1)=-3.5\] \[f(0)=0\] which of course means \[a_0=0\]
amistre64
  • amistre64
a quartic f(x) can have 3 zeros if its a touch and turn, coupled with a thru and thru
amistre64
  • amistre64
that would have 3 f' zeros then
anonymous
  • anonymous
it 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 \[f(x)=a(x+2)(x-4)x\]
amistre64
  • amistre64
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anonymous
  • anonymous
ho ho , that is a 4th degree poly. this one is sposed to be cubic!
amistre64
  • amistre64
how does a cubic have 3 zeros in the derivative? unless the original question is in error....
anonymous
  • anonymous
the derivative only has two zeros: -1 and 2
amistre64
  • amistre64
"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
  • anonymous
it is a parabola facing down, negative until -1, positive between -1 and 2, then negative
amistre64
  • amistre64
f' = 0 at -1,-1,5 is my best interp of it.... right or wrong lol
anonymous
  • anonymous
no it is not error. what it means is that f'(-1) = 0 and f(-1)=-3.5
amistre64
  • amistre64
ooohhh....... that helps, thnx :)
anonymous
  • anonymous
now let me see if i can finish this. before the day is over. lol
anonymous
  • anonymous
ok where were we before we were interrupted?
anonymous
  • anonymous
oh yes, you have a cubic polynomial with zeros at -2,0, and 4 so it must look like \[f(x)=a(x+2)(x-4)x\] 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
  • anonymous
The question says that f'(x) is 0 at -1,-3 and 5. But f'(-1)=0 yields a=0!
anonymous
  • anonymous
i guess i cannot read. i read it as \[f'(-1)=0\] at \[(-1,-3.5)\]
anonymous
  • anonymous
Yes but \[f^\prime(-1)=0 \implies a=0\]
anonymous
  • anonymous
actually i get that too. damn
anonymous
  • anonymous
the 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
anonymous
  • anonymous
in fact it get \[-6a=-8a\]
anonymous
  • anonymous
@satellite73 that means a=0.
anonymous
  • anonymous
think too many constraint to solve this one. yes, know it means a=0 that was what was bothering me.
amistre64
  • amistre64
if a = 0, just use a different letter lol
anonymous
  • anonymous
lets go slow. do we agree that \[f(x)=a(x+2)(x-4)x\] ?
anonymous
  • anonymous
because f is cubic and the roots are given.
anonymous
  • anonymous
The cubic is \[f(x)=ax(x+2)(x-4)\].
anonymous
  • anonymous
ok so far so good. and we agree that a < 0 yes?
anonymous
  • anonymous
thx a mil for the awesome insight, any chance i would be able to graph this equation.....
anonymous
  • anonymous
why?
anonymous
  • anonymous
because of the behavior of the derivative: negative then positive then negative means f is decreasing then increasing then decreasing. so a < 0
anonymous
  • anonymous
to show max an min values and intercepts
anonymous
  • anonymous
and we also agree that \[f(x)=ax^3-2ax^2-8ax\]
anonymous
  • anonymous
i 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.
anonymous
  • anonymous
-4 that is
anonymous
  • anonymous
no it says 4
anonymous
  • anonymous
f(4)=0
amistre64
  • amistre64
when 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
  • amistre64
little backwards, but still i quandry
anonymous
  • anonymous
I think -1,-3.5 and the f'(x) being 0 date are wrong.
anonymous
  • anonymous
it says f'(-1)=0 it changes direction at (-1,-3.5)
amistre64
  • amistre64
f'(x)=0 could also indicate an inflection point... right?
anonymous
  • anonymous
it 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
  • anonymous
no it is not an inflection point because the derivative changes sign there.
amistre64
  • amistre64
if the slope is the same on both sides of f' = 0, that indicates inflection i believe
amistre64
  • amistre64
the signs dont change at -3.5
anonymous
  • anonymous
derivative changes sign. that is what it says.
anonymous
  • anonymous
is inflexion point noy determind by f''?
amistre64
  • amistre64
in inflection point is made sure by f''; but an inflection point can still have a 0 slope in f'
anonymous
  • anonymous
says f'(x)<0 for x<-1 f'(x)>0 for -12
anonymous
  • anonymous
derivative changes sign so f changes direction from decreasing to increasing. still i cannot come up with a. will try later.
amistre64
  • amistre64
"when f'(x) = 0 at (-1;-3.5) then f'(x)<0 for x<-1 and x>2 f'(x)>0 for -1
anonymous
  • anonymous
-3.5 is an inflection point..
anonymous
  • anonymous
thx for all the help, egan helmie, south africa grade 12 -2011-provincial assesment task
amistre64
  • amistre64
I see f(x) looking somthing like this if i interpret your info correctly
1 Attachment
anonymous
  • anonymous
oh 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
  • anonymous
in 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 \[f(x)=ax(x+2)(x-4)=ax^3-2ax^2-8ax\] thus \[f'(x)=3ax^2-4ax-8a\] and \[f'(-1)=3a+4a-8a\] the only way for this to be 0 is for \[a=0\] which is a contradiction.
amistre64
  • amistre64
i trust your interpretations and its method; I dont trust that the given information is rigid enough to come to a solid conclusion tho ;)
anonymous
  • anonymous
it 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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