You need to evaluate the derivative of the function at `x = 3` , using the difference quotient definition, such that:

`lim_(x->3) (f(x) - f(3))/(x - 3) = f'(3)`

`lim_(x->3) (x^2 - 2x - 2 - (3^2 - 6 - 2))/(x - 3) = lim_(x->3) (x^2 - 2x - 2...

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You need to evaluate the derivative of the function at `x = 3` , using the difference quotient definition, such that:

`lim_(x->3) (f(x) - f(3))/(x - 3) = f'(3)`

`lim_(x->3) (x^2 - 2x - 2 - (3^2 - 6 - 2))/(x - 3) = lim_(x->3) (x^2 - 2x - 2 - 9 + 6 + 2)/(x - 3)`

`lim_(x->3) (f(x) - f(3))/(x - 3) = lim_(x->3) (x^2 - 2x - 3)/(x - 3) = 0/0`

Since 3 is a solution to the quadratic polynomial `x^2 - 2x - 3` , you may write its factored form, such that:

`lim_(x->3) (x^2 - 2x - 3)/(x - 3) = lim_(x->3) ((x - 3)(x + 1))/(x - 3)`

Reducing duplicate factors yields:

`lim_(x->3) ((x - 3)(x + 1))/(x - 3) = lim_(x->3) (x + 1) = 3 + 1 = 4`

**Hence, evaluating the derivative of the given function, at the point ` x = 3` , using quotient definition, yields `f'(3) = 4.` **