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
- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

use cardano's method
given below

##### 1 Attachment

- 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...

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## 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

##### 1 Attachment

- 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

##### 1 Attachment

- 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.