Use the four-step process to find $$ f^{\prime}(x) $$ and then find $$ f^{\prime}(-1), f^{\prime}(1) $$, and $$ f^{\prime}(7) $$. $$ f(x)=2 x^{2}-x-3 $$ $$ f^{\prime}(x)=\square $$

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Find the first order derivative with respect to $$ x $$ for $$ 2x^{2}-x-3 $$.
Evaluate the derivative by following steps:
- step0: Evaluate the derivative:
  $$\frac{d}{dx}\left(2x^{2}-x-3\right)$$
- step1: Use differentiation rules:
  $$\frac{d}{dx}\left(2x^{2}\right)-\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left(3\right)$$
- step2: Calculate:
  $$4x-1-0$$
- step3: Remove 0:
  $$4x-1$$
The first derivative of the function $$ f(x) = 2x^{2} - x - 3 $$ is $$ f'(x) = 4x - 1 $$.

Now, we need to find the values of $$ f'(-1), f'(1), $$ and $$ f'(7) $$ by substituting the respective values of $$ x $$ into the derivative function $$ f'(x) = 4x - 1 $$.

1. $$ f'(-1) = 4(-1) - 1 = -4 - 1 = -5 $$
2. $$ f'(1) = 4(1) - 1 = 4 - 1 = 3 $$
3. $$ f'(7) = 4(7) - 1 = 28 - 1 = 27 $$

Therefore, $$ f'(-1) = -5 $$, $$ f'(1) = 3 $$, and $$ f'(7) = 27 $$.
$$ f'(x) = 4x - 1 $$ $$ f'(-1) = -5 $$, $$ f'(1) = 3 $$, $$ f'(7) = 27 $$

Use the four-step process to find and then find , and .

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