Question

Give an example of a function, g(x), that has a local maximum
at (-3; 3) and a local minimum at (3, -3).

Answer 1

g'(3) = -3 eh...

Answer 2

its a cubic at its basic

Answer 3

centered at the the origin

Answer 4

g'(3) = 0

and g'(-3) = 0

Answer 5

x^3 derived is 3x^2

3(3)^2 = 0

27 = 0

27 - 27 = 0

Answer 6

3x^2 - 27 = g'

x^3 - 27x +C = function of g(x); just gotta calibrate for (3,-3)

Answer 7

-3 = (3)^3 -27(3) + C

-3 -17 + 27(3) = C

C = 61


g(x) = x^2 -27x + 61 perhaps?

Answer 8

g(x) = x^3 -27x + 61 that is lol

Answer 9

Thanks a lot. I really appreciate it.

Answer 10

it might be wrong; i havent checked it good wnough yet :)

Answer 11

C = 51 maybe...

Answer 12

Hmm, when I graph it it the maximum and minimum isn't (-3, 3) and (3, -3) let me try with C=51

Answer 13

i see it too; :) gotta figure out where I looked at it wrong

Answer 14

g' = 3x^2 - 27 = 0 at -3 and 3 so the derivative is good

Answer 15

g(x) = x^3 - 27x + C if we integrate it...

Answer 16

(x=-3,y=3) should fit it and (x=3,y=-3) should fit it, so we solve for C with these parameters

Answer 17

-3 = (3)^3 - 27(3) + C and 3 = (-3)^3 - 27(-3) + C


-3 - (3)^3 + 27(3) = C OR 3 - (-3)^3 + 27(-3) = C

Answer 18

C = 51 or C = -51 which doesnt work out to well in teh end

Answer 19

g(0) would also have to equal 0

Answer 20

C = 0

Answer 21

I just graphed it again and the minimum is at (3, -3) but not the Maximum

Answer 22

the maximum is not at (3, -3)

Answer 23

g(x) = x^3 - 27x ... thats it

Answer 24

ack!!... that aint it either.. lol

Answer 25

soooo close

Answer 26

Haha, it's ok, thanks a lot for the help though. But I got a little help from someone but have no idea how hey got it, maybe you can help elaborate this.
For this problem they started from this:
(x+3)(x-3)
g'(x)= X^2 -9
= 1/3x^3 -9

???? Have no idea as to how they got here for this problem.

Answer 27

i got some idea, but have to look at it some more

Answer 28

g'(x) has roots at 3 and -3 so

g'(x) = x^2 - 9 if we combine them; which is good; then integrate up to

x^3
--- - 9x tho
3

Answer 29

(1/3)(x^3) - 9x + C is the family of curves for it....

Answer 30

C = -15 then i think

(1/3)x^3 - 9x -15 ; lets check it

Answer 31

Now the max is right but not the min...

Answer 32

I did this very quickly; \(f(x)=x^3-27x-51\). Check it!!

Answer 33

did that first lol

Answer 34

it works for one , but not the other

Answer 35

g(0) = 0 has to be part of te graph

Answer 36

so adding somethig to it aint gonna help; we gotta find the right 'slope'

Answer 37

M(x^3 - Bx) is what we need

Answer 38

Dx(Mx^3 - MBx)

3Mx^2 - MB

M(3x^2 - B) = 0 when x = 3 or -3

Answer 39

Maybe include more powers of x?

Answer 40

or derive again for inflection...

Answer 41

6Mx = 0 when x = 0 lol ... that dint work out to well

Answer 42

haha, thanks, i really appreciate the help

Answer 43

k(-3)^3 = 3

-27k = 3

k = -1/9 maybe?

x^3
---- ; 3x^2/9 = x^2/3 i wonder
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