find the relative max and min
f(x)=(x+8)^2(3-x)

Question

find the relative max and min 
f(x)=(x+8)^2(3-x)

anonymous · Answer

i found the relative min which is -8 but am not sure about the relative max

anonymous · Answer

did you take the derivative, set it equal to zero and solve?

anonymous · Answer

this is a cubic polynomial with negative leading coefficient, a zero at  $-8$ with multiplcity 2, and a zero at $3 $

anonymous · Answer

thats how i found -8 but i also thought 3 would be an answer but it isnt so im not quite sure what to do

anonymous · Answer

you need the derivative
did you find that?

anonymous · Answer

i got (x+8)^2 (-1) + (2) (x+8) (3-x)

anonymous · Answer

you need the zeros, so you actually have to multiply out and combine like terms to solve the quadratic equation 
you should get 
$$-3x^2-26x-16$$ set it equal to zero and solve]
zeros are $-8$ which is the $x$ coordinate of the local min, (the $y$ coordinate is obliviously 0) and $-\frac{2}{3}$ which is the $x$ coordinate of the local max

anonymous · Answer

oooo i see...so you have to multiply out each one out and then add together like terms?

anonymous · Answer

thanks! can you help me with another one please :):)