First, find the derivative of $$f(x)$$: $$f'(x) = \frac{1}{4}(4x^3) - \frac{1}{2}(2x) + 0 = x^3 - x$$. Now, set the derivative equal to zero: $$x^3 - x = 0$$. Factor out an $$x$$: $$x(x^2 - 1) = 0$$. This gives solutions $$x=0$$ and $$x^2=1$$, so $$x = -1, 0, 1$$. From the given choices, $$x=1$$ is a correct answer.

To find the derivative of $$f(x)$$, we apply the power rule, sum/difference rule, and constant rule to each term. The derivative of $$5x^3$$ is $$15x^2$$. The derivative of $$-2x^2$$ is $$-4x$$. The derivative of $$x$$ is $$1$$. The derivative of the constant $$-9$$ is $$0$$. Combining these gives $$f'(x) = 15x^2 - 4x + 1$$.

The velocity function is the first derivative of the position function: $$v(t) = p'(t) = 3t^2 - 12t$$. The acceleration function is the second derivative of the position function: $$a(t) = p''(t) = 6t - 12$$. To find when the acceleration is zero, set $$a(t) = 0$$: $$6t - 12 = 0$$, which gives $$t = 2$$. To find the velocity at this time, substitute $$t=2$$ into the velocity function: $$v(2) = 3(2)^2 - 12(2) = 3(4) - 24 = 12 - 24 = -12$$.

First, expand the expression for $$f(x)$$: $$f(x) = (2x - 3)(2x - 3) = 4x^2 - 6x - 6x + 9 = 4x^2 - 12x + 9$$. Now, differentiate the expanded polynomial term by term: $$f'(x) = rac{d}{dx}(4x^2 - 12x + 9) = 8x - 12$$.

First, find the derivative of $$h(x)$$ using the sum, difference, and constant multiple rules. $$h'(x) = 4f'(x) - 3g'(x) + 0$$. Now, substitute $$x=2$$ into the expression for $$h'(x)$$: $$h'(2) = 4f'(2) - 3g'(2)$$. Using the given values, $$h'(2) = 4(5) - 3(-1) = 20 + 3 = 23$$.

Linearity rules allow us to handle the sum and differences in f(t) by differentiating individually with the constant multiple rule. For f(t) = $9t^3$ - $2t^2$ - 6t + 1, apply the power rule to every term. The derivative of $9t^3$ is $27t^2$, of $-2t^2$ is -4t, of -6t is -6, and of 1 is 0. Therefore, f'(t) = $27t^2$ - 4t - 6, corresponding to choice A. A common rule misuse is applying the power rule without multiplying by the exponent, e.g., $t^3$ to $t^2$ instead of $3t^2$. Another is ignoring negative signs in differences. A transferable strategy is to process each term with d/dx (c $x^n$) = c n $x^{n-1}$, remember constants vanish, and maintain sign integrity in the final expression.

Linearity rules allow independent differentiation of terms in F(x) = $3x^3$ - $4x^2$ + 5x - 6. Derivative of $3x^3$ is $9x^2$, of $-4x^2$ is -8x, of 5x is 5, of -6 is 0. Thus, F'(x) = $9x^2$ - 8x + 5. A common misuse is subtracting instead of adding derivatives in sums. Constants are often incorrectly included. Decompose using linearity and apply power rule to each term for accuracy.

The sum and difference rules of linearity allow term-by-term differentiation of polynomials. For G(x) = $-6x^2$ + 13x - 15, derivative of $-6x^2$ is -12x, of 13x is 13, and of -15 is 0. Thus, G'(x) = -12x + 13. A common misuse is including the constant's derivative as nonzero, like adding -15. Signs on linear terms must be preserved accurately. For quadratic functions, use linearity to differentiate each power separately, ensuring constants are ignored.

Linearity rules facilitate term-by-term differentiation for y(t) = $13t^2$ + 4t + 9. Derivative of $13t^2$ is 26t, of 4t is 4, of +9 is 0. Thus, y'(t) = 26t + 4. A common misuse is including the constant in the derivative, like adding 9. Linear terms are sometimes overlooked. For quadratics, apply sum rules and power rule individually to each term.

Linearity rules facilitate term-by-term differentiation for N(x) with sums and multiples. For N(x) = $5x^4$ + $7x^3$ - 14x + 9, use the power rule. The derivative of $5x^4$ is $20x^3$, of $7x^3$ is $21x^2$, of -14x is -14, and of 9 is 0. Therefore, N'(x) = $20x^3$ + $21x^2$ - 14, which is choice A. A common misuse is forgetting multiplication for cubic terms. Some include constants. A transferable strategy is to sequence terms, apply d/dx (c $x^n$) = c n $x^{n-1}$, and combine using linearity.