Help me simplify this down some more so I can set it to zero and find my critical points.
f(x) = x^3 (x+2)^6
f'(x) = x^3 * 6(x+2)^5 + 3x^2 * (x+2)^6
I see a common factor of 3x^2(x+2)^5
so
x^3 * 6(x+2)^5 + 3x^2 * (x+2)^6
= 3x^2(x+2)^5[2x + x+2}
= 3x^2(x+2)^5(3x+2)
Thanks.
To find the critical points of a function, you need to first find its derivative and set it to zero. In this case, you have already found the derivative of the function f(x) = x^3 (x+2)^6, which is f'(x) = x^3 * 6(x+2)^5 + 3x^2 * (x+2)^6.
To simplify further, let's factor out the common terms.
f'(x) = x^3 * 6(x+2)^5 + 3x^2 * (x+2)^6
= 6x^3(x+2)^5 + 3x^2(x+2)^6
Now we can see that both terms have a common factor of (x+2)^5. Let's factor that out.
f'(x) = (x+2)^5(6x^3 + 3x^2(x+2))
Looking at the second term, we can simplify it by combining the like terms inside the parentheses.
f'(x) = (x+2)^5(6x^3 + 3x^3 + 6x^2)
= (x+2)^5(9x^3 + 6x^2)
Now that the derivative has been simplified, we can proceed to finding the critical points by setting f'(x) to zero:
(x+2)^5(9x^3 + 6x^2) = 0
To find the critical points, we need to set each factor to zero separately and solve for x.
Setting (x+2)^5 = 0 gives us x = -2, since a non-zero number raised to the power of 5 can never be zero.
Next, setting 9x^3 + 6x^2 = 0, we can factor out x^2.
x^2(9x + 6) = 0
Setting each factor to zero separately, we get x = 0 and x = -2/3.
So, the critical points of the function f(x) = x^3 (x+2)^6 are x = -2, x = 0, and x = -2/3.