Question

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 14 years ago

Answer 1

use cardano's method
given below

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

good morning.

Answer 7

howdy :)

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

that would have 3 f' zeros then

Answer 15

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
\[f(x)=a(x+2)(x-4)x\]

Answer 16

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

Answer 17

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

Answer 18

the derivative only has two zeros: -1 and 2

Answer 19

"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...

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

ok where were we before we were interrupted?

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

actually i get that too. damn

Answer 31

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<x<2
f(0) =0
f(-2)=0 and f(4)=0

Answer 32

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

Answer 33

@satellite73 that means a=0.

Answer 34

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

Answer 35

if a = 0, just use a different letter lol

Answer 36

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

Answer 37

because f is cubic and the roots are given.

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

why?

Answer 42

because of the behavior of the derivative: negative then positive then negative means f is decreasing then increasing then decreasing. so a < 0

Answer 43

to show max an min values and intercepts

Answer 44

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

Answer 45

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.

Answer 46

-4 that is

Answer 47

no it says 4

Answer 48

f(4)=0

Answer 49

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 +

????????

Answer 50

little backwards, but still i quandry

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

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.

Answer 55

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

Answer 56

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

Answer 57

the signs dont change at -3.5

Answer 58

derivative changes sign. that is what it says.

Answer 59

is inflexion point noy determind by f''?

Answer 60

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

Answer 61

says f'(x)<0 for x<-1
f'(x)>0 for -1<x<2
f"(x)< 0 for x>2

Answer 62

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

Answer 63

"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"

Answer 64

-3.5 is an inflection point..

Answer 65

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

Answer 66

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

Answer 67

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.

Answer 68

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.

Answer 69

i trust your interpretations and its method; I dont trust that the given information is rigid enough to come to a solid conclusion tho ;)

Answer 70

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.