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

- anonymous

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- amistre64

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

- amistre64

its a cubic at its basic

- amistre64

centered at the the origin

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## More answers

- amistre64

g'(3) = 0
and g'(-3) = 0

- amistre64

x^3 derived is 3x^2
3(3)^2 = 0
27 = 0
27 - 27 = 0

- amistre64

3x^2 - 27 = g'
x^3 - 27x +C = function of g(x); just gotta calibrate for (3,-3)

- amistre64

-3 = (3)^3 -27(3) + C
-3 -17 + 27(3) = C
C = 61
g(x) = x^2 -27x + 61 perhaps?

- amistre64

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

- anonymous

Thanks a lot. I really appreciate it.

- amistre64

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

- amistre64

C = 51 maybe...

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

-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

- amistre64

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

- amistre64

g(0) would also have to equal 0

- amistre64

C = 0

- anonymous

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

- anonymous

the maximum is not at (3, -3)

- amistre64

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

- amistre64

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

- amistre64

soooo close

- anonymous

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.

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

C = -15 then i think
(1/3)x^3 - 9x -15 ; lets check it

- anonymous

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

- anonymous

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

- amistre64

did that first lol

- amistre64

it works for one , but not the other

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

Dx(Mx^3 - MBx)
3Mx^2 - MB
M(3x^2 - B) = 0 when x = 3 or -3

- amistre64

Maybe include more powers of x?

- amistre64

or derive again for inflection...

- amistre64

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

- anonymous

haha, thanks, i really appreciate the help

- amistre64

k(-3)^3 = 3
-27k = 3
k = -1/9 maybe?
x^3
---- ; 3x^2/9 = x^2/3 i wonder
9

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