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OpenStudy (anonymous):
Wat is the derivative of g(x)=2x^3-6x^2-18x
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OpenStudy (anonymous):
Use the power rule on each individual term in the function:
\[\frac{d(a*x^n)}{dx} = a * n * x^{n-1}\]
where "a" is a constant.
OpenStudy (anonymous):
Please teach me how to do this
OpenStudy (anonymous):
g'(x) = 6x^2 - 12x - 18
OpenStudy (anonymous):
Thanks alot
OpenStudy (anonymous):
how do you find the critical points?
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OpenStudy (anonymous):
Set the derivative equal to zero and solve for x, and you'll get your critical points.
OpenStudy (anonymous):
can you show me please?
OpenStudy (anonymous):
Just set g'(x) = 0, solve the quadratic (it factors pretty cleanly, I think) and the values of x that come out are critical points.
OpenStudy (anonymous):
So set it equal to 0 and I get 0= 6x^2-12x-18 and i can factor that to 0=6(x^2-2x-3) then what do i do?
OpenStudy (anonymous):
Eliminate the 6, and factor x^2-2x-3. That's something very basic for calc, you should be able to do it. (x-3)(x+1).
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OpenStudy (anonymous):
so 3 and -1 are the critical points?
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