the derivative of an equation is 4x(x^2-4) it wants me to find the location of relative maximum, minimum and inflection points how do I do this?

Question

mrnood · Answer

those points all are found when the derivative =0

so you need the followowing steps to start

first factorise the term in brackets so you have
(4x)(x-a)(x-b)

freckles · Answer

so we have 
$$y'=4x(x^2-4) \text{ for some } y=f(x)$$
We can find the critical points f by setting y'=0

anonymous · Answer

they are 0 and 2 right

freckles · Answer

$$\text{ and } x=-2$$

mrnood · Answer

no - there are 3 points - just factorise as I said first

mrnood · Answer

or just let freckles post the answer for you

anonymous · Answer

2,0, and -2 okay then I do a sin chart?

freckles · Answer

well he had part of the answer

freckles · Answer

I think since it is a cal question he knows how to solve x^2-4=0

freckles · Answer

he just forgot
or i could be wrong

mrnood · Answer

the way I would approach it is then to take the second derivative - i.e. differentiat the first derivative ....

mrnood · Answer

d (uv) = vdu + udv
 dx           dx     dx

mrnood · Answer

If the second derivative is -ve at the critical point then the point is a maximum
If the second derivative is +ve at the critical point then the point is a minimum
If the second derivative is 0 at the critical point then the point is a POI

anonymous · Answer

so I get the answer by taking the second derivative?

mrnood · Answer

take the second derivative f''

For each of the three x values you got above (0,2,-2) put them into the formula for the f'' and evaluate the value of f''

Then apply the rules in my last post to decide whether the point is a max, min or POI