How many extrema (maxima and minima) does the function f(x) = (x+2)^3 * (x-5)^2 have? PS...no calculator allowed for this question or else i would've just graphed it
Take the derivative and set it equal to zero, take that answer and plug it into the second derivative, if it is negative, then it is a maxima and if it is positive, then it is a minima.
im not sure how to differentaite that
Product rule. 3(x+2)^2 * (x-5)^2 + 2(x-5)*(x+2)^3
how wud i possibly solve taht for 0?
can that be simplified or smthng
Ha yeah I guess that seems a bit tough. If you have a calculator you can type it in and find where y=0 to find the answers. If not then you will have to do it the long way.
Ok I think you need to be able to see this typed as an equation without the little "^" signs... http://www.wolframalpha.com/input/?i=3%28x%2B2%29^2+*+%28x-5%29^2+%2B+2%28x-5%29*%28x%2B2%29^3%3D0 you ca factor out (x+2)^2 and (x-5) which should make the equation a lot easier to solve.
im not sure how it simplified into (x-5)*(x+2)^2 * (5x-11)...idk where the 3rd part comes from
Do you remember the product rule?
yeah f'g +fg' / g^2
Wait, where did you get (5x-11) from??
tahts wat wolframalpha says
No, that's the quotient rule, that would be if you are dividing, well no actually you would need to have f'g - fg' / g^2 for the quotient rule. The product rule is just f'g +fg'
is an alternate form of it...im assuming tahts wat it is simpleied
Oh, no you are looking at the "alternate forms" that is because it is simplified. Just look at the main equation at the top.
oh ok so x=5 or x = -2 then
Yes!
but when i do s sign diagram i get only one maxima but the asnwer is 2 maximas
Never mind i get it wen i llok at the graph.
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