Give the value of x where F(x) = X^3-33/2X+84x-2 what is the local minimum?
how do i find the roots of this?
you mean the roots of F or the roots of \(F'\)?
oh yes
if you are looking for a local max or mean you need the roots of \(F'\) ok we need to find \(F'\) first
i got 3X^2 -66/2x+84=0
i am not to good with finding roots in general so many different methods and coeffecients mess me up.Can you give me general way that will work for all types of polynomials?
no and it looks like your derivative is not right. here is what wolfram told me http://www.wolframalpha.com/input/?i=%28x^3-33%29%2F%282x%2B84x-2%29
you have to use the quotient rule when taking the derivative. it looks like you took the derivative top and bottom separately, and that doesn't work
im sorry its 33/2x ^2 seperate from rest
\[x^3-\frac{33}{2x}+84x-2 \]
made mistakae here is the correct f(x)= x^3-33/2x^2 +84x-2
did i get it right?
i fixed it
do you see it?
yeah it is this i think \[x^3-\frac{33}{2x^2}+84x-2\]
x^2 comes after 33/2
oooooooooooooooooooooh
lol im sorry for the trouble
\[x^3-\frac{33}{2}x^2+84x-2\] in that case we are in good shape no trouble take the derivative get a polynomial of degree 2 if it does not factor, use the quadratic formula!
\[f'(x)=3x^2-33x+84\] set the equal zero and solve
is quadratic formula practical on a test?
bet this one factors \[f'(x)=3(x^2-11x+28)\]
maybe we get lucky \[x^2-11x+28=0\] \[(x-7)(x-4)=0\] etc
nice!
so we get \(x=4, x=7\) as the critical points, obviously 4 will give a local max and 7 a local min
so with those points u plugged back into f'(x) or F(x)?
oh be careful here
if you plug them back in to \(f'\) you will get 0 because that is how we found them to begin with if you want the y values plug them in to \(f\) not \(f'\)
thanks
yw
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