High Degree Polynomial Turning Points Help

[AKYBY2TMA]

I have this function which I need help finding all of the turning points:

f(x) = (x+1)^(2)(x-5)^(3)(x+4)^(4)(x-3)^(5)

Can anyone please help me solve it.
If some rules or theorem are using please list them.
And most importantly, please include some explanation and full working out of how you found the answer. Please, that would be really helpful.

Thanks.

 

[daon2]

Turning points will occur at values c such that f'(c)=0. So to find all possible values, take the derivative, set it equal to 0, and solve for x. (If applicable, make sure to exclude 2-dimensional "saddle points")

The derivative of that looks a little nasty from a product-rule standpoint. However, if c is a multiple root of f(x) then f'(c)=0. NOTE:All of the roots of f(x) are multiple roots!

 

[HallsofIvy]

I have this function which I need help finding all of the turning points:

f(x) = (x+1)^(2)(x-5)^(3)(x+4)^(4)(x-3)^(5)

Differentiate usig the product rule:
f'= 2(x+1)(x-5)^(3)(x+4)^(4)(x-3)^(5)+ 3(x+1)^(2)(x-5)^(2)(x+4)^(4)(x-3)^(5)+ 4(x+1)^(2)(x-5)^(3)(x+4)^(3)(x-3)^(5)+ 5(x+1)^(2)(x-5)^(3)(x+4)^(4)(x-3)^4

Setting that equal to 0, we are helped by the fact that there are so many multpiple factors:
(x+ 1)(x- 5)^2(x+ 4)^3(x- 3)^4(2(x- 5)(x+4)(x- 3)+ 3(x+ 1)(x+ 4)(x- 3)+ 4(x+ 1)(x- 5)(x- 3)+ 5(x+ 1)(x- 5)(x+ 4))=0
so that x= -1, x= 5, x=- - 4 and x= 3 are possible turning points. But for a true "turning point", the derivative must actually change sign and that only happens when the right factor has odd power. So the turning points, so far, are x= -1 and x= -4. We still would need to find x where 2(x- 5)(x+4)(x- 3)+ 3(x+ 1)(x+ 4)(x- 3)+ 4(x+ 1)(x- 5)(x- 3)+ 5(x+ 1)(x- 5)(x+ 4)= 0.

Can anyone please help me solve it.
If some rules or theorem are using please list them.
And most importantly, please include some explanation and full working out of how you found the answer. Please, that would be really helpful.

Thanks.