Question

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

Answer 1

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.

Answer 2

Please teach me how to do this

Answer 3

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

Answer 4

Thanks alot

Answer 5

how do you find the critical points?

Answer 6

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

Answer 7

can you show me please?

Answer 8

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.

Answer 9

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?

Answer 10

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).

Answer 11

so 3 and -1 are the critical points?