Where r the turning points of x^2(x-4)^7(x+3)^6

There are 5  turning points at x = -3, -1.11, 0, 1.44 and 4.

Step-by-step explanation:

The product rule for 3 terms is

F' (x) =   f'(x) g(x) h(x) + f(x) g'(x) h(x) + f(x) g(x) h'(x)

F(x) = x^2(x - 4)^7 (x + 3)^6

F' (x) =  2x(x-4)^7(x+3)^6 + x^2 7(x - 4)^6 (x + 3)^6 + x^2(x-4)^7 6(x+3)^5 = 0 at the turning points.

That is some equation!!

By observation, 3 roots are x = 0 , x = 4 and  x = -3 so there are turning points at these values of x.

I'll use  software to graph  it to find any more real solutions - they are -1.11 and 1.44 to nearest hundredth.

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.

i do not know the answer

step-by-step explanation: