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
- 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
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- schrodinger

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- anonymous

use cardano's method
given below

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- 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

is the derivative the cubic? it has 3 zeros...

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## More answers

- amistre64

f' = 0 at -1,-3,5 ?

- anonymous

\[f'(x)=ax^2-ax-2a\] and therefore
\[f(x) = \frac{ax^3}{3}-\frac{ax^2}{2}-2ax+c\]

- anonymous

good morning.

- amistre64

howdy :)

- anonymous

i assume this is looking for the cubic so the derivative is quadratic. with zeros at -1 and 2

- amistre64

"please help to find cubic equation when f' (x) =0 at(-1;-3,5)" ??

- anonymous

i am assuming you are looking for the cubic function whose derivative is zero at -1

- anonymous

in other words \[f(x)=a_3x^3+a_2x^2+a_1x+a_0\]

- anonymous

\[f(-1)=-3.5\]
\[f(0)=0\]
which of course means \[a_0=0\]

- amistre64

a quartic f(x) can have 3 zeros if its a touch and turn, coupled with a thru and thru

- amistre64

that would have 3 f' zeros then

- 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

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- anonymous

ho ho , that is a 4th degree poly. this one is sposed to be cubic!

- amistre64

how does a cubic have 3 zeros in the derivative? unless the original question is in error....

- anonymous

the derivative only has two zeros: -1 and 2

- 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

it is a parabola facing down, negative until -1, positive between -1 and 2, then negative

- amistre64

f' = 0 at -1,-1,5 is my best interp of it.... right or wrong lol

- anonymous

no it is not error. what it means is that f'(-1) = 0 and f(-1)=-3.5

- amistre64

ooohhh....... that helps, thnx :)

- anonymous

now let me see if i can finish this. before the day is over. lol

- anonymous

ok where were we before we were interrupted?

- 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

The question says that f'(x) is 0 at -1,-3 and 5. But f'(-1)=0 yields a=0!

- anonymous

i guess i cannot read. i read it as
\[f'(-1)=0\] at \[(-1,-3.5)\]

- anonymous

Yes but \[f^\prime(-1)=0 \implies a=0\]

- anonymous

actually i get that too. damn

- 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

in fact it get \[-6a=-8a\]

- anonymous

@satellite73 that means a=0.

- anonymous

think too many constraint to solve this one. yes, know it means a=0 that was what was bothering me.

- amistre64

if a = 0, just use a different letter lol

- anonymous

lets go slow. do we agree that \[f(x)=a(x+2)(x-4)x\] ?

- anonymous

because f is cubic and the roots are given.

- anonymous

The cubic is \[f(x)=ax(x+2)(x-4)\].

- anonymous

ok so far so good. and we agree that a < 0 yes?

- anonymous

thx a mil for the awesome insight, any chance i would be able to graph this equation.....

- anonymous

why?

- 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

to show max an min values and intercepts

- anonymous

and we also agree that
\[f(x)=ax^3-2ax^2-8ax\]

- 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

-4 that is

- anonymous

no it says 4

- anonymous

f(4)=0

- 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

little backwards, but still i quandry

- anonymous

I think -1,-3.5 and the f'(x) being 0 date are wrong.

- anonymous

it says f'(-1)=0 it changes direction at (-1,-3.5)

- amistre64

f'(x)=0 could also indicate an inflection point... right?

- 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

no it is not an inflection point because the derivative changes sign there.

- amistre64

if the slope is the same on both sides of f' = 0, that indicates inflection i believe

- amistre64

the signs dont change at -3.5

- anonymous

derivative changes sign. that is what it says.

- anonymous

is inflexion point noy determind by f''?

- amistre64

in inflection point is made sure by f''; but an inflection point can still have a 0 slope in f'

- anonymous

says f'(x)<0 for x<-1
f'(x)>0 for -12

- anonymous

derivative changes sign so f changes direction from decreasing to increasing. still i cannot come up with a. will try later.

- 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

-3.5 is an inflection point..

- anonymous

thx for all the help, egan helmie, south africa grade 12 -2011-provincial assesment task

- amistre64

I see f(x) looking somthing like this if i interpret your info correctly

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- 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

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

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

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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