Another method to calculate a derivative of a function in a point is:

f'(1) = limit [f(x)-f(1)]/(x-1) = lim (x^3 + x^2 -2x - 0)/(x-1)=0/0

x->1 x->1

It is obvious that x=1 is a root for f(x)=x^3 + x^2 -2x, so the second root is:

x1 + x2 = -1

1 + x2 = -1

x2 = -2

We'll write f(x) as a product of factors:

f(x) = (x-1)(x+2)

Now, we'll calculate the limit:

**f'(1) = lim (x-1)(x+2)/(x-1) = lim (x+2) = 1+2 = 3**

**f'(-1)=lim [f(x)-f(-1)]/(x+1)=lim (x^3 + x^2 -2x+2)/(x+1)=-1**

f'(-1) = 3(-1)^2 + 2(-1) - 2 = 3-4 = -1

The sum:

**f'(1) + f'(-1) = 3-1 = 2**

Given f(x) =x^3+x^2-2x. To find the value of f'(1)+f'(-1).

Solution:

f(x) = x^3+x^2-2x

We find the differential coefficient f'(x) of f(x) and then find the values of f'(x) at x=1 and x=-1 and then find the sum of them, that is f'(1) +f'(-1).

f'(x) = (d/dx) (x^3+x^2-x)

= (d/dx)(x^3)+(d/dx)(x^2)+(d/dx)(-2x). We use (d/dx)(kx^n) = knx^(n-1).

=3x^(3-1)+2x^(2-1)+(-2*1x^(1-1)

=3x^2+2x-2

f'(1) = 3*1^2+2*1-2 = 3

f'(-1) = 3*(-1)^2+2*(-1)-2 =3-2-2 =-1.

Therefore the value of f'(1)+f(-1) = 3+(-1) = 2

Given:

f(x) = x^3 + x^2 - 2x

Therefore:

f'(x) = 3x^2 + 2x - 2

f'(1) = 3(1)^2 + 2(1) - 2

= 3 + 2 - 2 = 3

f' (-1) = 3(-1)^2 + 2(-1) - 2

= 3 - 2 - 2 = -1

Therefore:

f'(1) + f'(-1) = 3 - 1 = 2