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Mathematics 13 Online
OpenStudy (anonymous):

Wat is the derivative of g(x)=2x^3-6x^2-18x

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?

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).

OpenStudy (anonymous):

so 3 and -1 are the critical points?

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