The graph of the function has more than one turning point. Use the graphing calculator to find the coordinates of the turning point that lies in the given interval. f(x)= 7x^3 -9x^2 -x+1, [-1, 10] The coordinates of the turning point are
Is this calculus?
College Algebra
Find f'(x)
The asker said algebra, so presumably that means no calculus allowed.
So what do I do?
it would certainly be a much easier calculus problem not sure I know a method for this using only algebra, I'll have to think on it...
ok
Graph it
that's all I can think of too
And the calculus way?
find x=a such that f'(a)=0
...in the given interval
http://www.wolframalpha.com/input/?i=plot+y%3D7x^3+-9x^2+-x%2B1
ok so what would the coordinates of the turning point?
It says "Use the graphing calculator to find the coordinates of the turning point that lies in the given interval"
wow I managed to not even read the post :P thanks cinar
I dont have one
(:
the website wolfram alpha is a pretty good one http://www.wolframalpha.com/
that site serves as a graphing calculator
laptop blocked it
http://answers.yahoo.com/question/index?qid=20091030113220AAieoZy
that method again requires calculus your three options are to buy the graphing calculator, get wolfram working on your laptop, or learn calculus besides that I have no ideas
I guess you have to use derivative
they are roughly (0.9,-2) (0.1,1) but exactly
It was (-0.05, 1.03)
http://www.wolframalpha.com/input/?i=max+y%3D7x^3+-9x^2+-x%2B1
wolfram disagrees o_O
http://www.wolframalpha.com/input/?i=min+y%3D7x^3+-9x^2+-x%2B1
that's right, I just used my eyes (:
(-0.05,1.03) and (0.91,-2.09) these are your answer..
I made typo for this they are roughly (0.9,-2) (0.1,1) but exactly
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