Answer:
The derivative of that product of functions evaluated at the point x=-2 gives "-3"
Step-by-step explanation:
Recall that the derivative of a product of two functions f(x) and g(x) is given by the formula:
(f*g)'= f' * g+ f * g'[tex](2(-2)+3)\,(-(-2)^2-3(-2)-2)+((-2)^2+3(-2)-1)\,(-2(-2)-3)=[/tex]
So it would be convenient to reduce this product of three functions to a product of just 2, performing (-x-1)*(x+2) = - x^2 - 2x - x -2 = - x^2 -3x -2
therefore we need to find the derivative of x^2 + 3x -1, and the derivative of - x^2 -3x -2 to obtain the answer:
[tex](x^2 + 3x -1)'=2x+3\\ \\(- x^2- 3x -2)'=-2x-3[/tex]
Now, applying the product rule for those two trinomial functions, we get:
[tex](2x+3)\,(-x^2-3x-2)+(x^2+3x-1)\,(-2x-3)[/tex]
which at x = -2 becomes:
[tex](2x+3)\,(-x^2-3x-2)+(x^2+3x-1)\,(-2x-3)\\(2(-2)+3)\,(-(-2)^2-3(-2)-2)+((-2)^2+3(-2)-1)\,(-2(-2)-3)= -3[/tex]
