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## anonymous 5 years ago Wat is the derivative of g(x)=2x^3-6x^2-18x

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1. 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.

2. anonymous

Please teach me how to do this

3. anonymous

g'(x) = 6x^2 - 12x - 18

4. anonymous

Thanks alot

5. anonymous

how do you find the critical points?

6. anonymous

Set the derivative equal to zero and solve for x, and you'll get your critical points.

7. anonymous

can you show me please?

8. 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.

9. 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?

10. 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).

11. anonymous

so 3 and -1 are the critical points?

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