h(x)=\\frac{x^{2}-9x + 18}{x - 4}\nhow would you rewrite h(x) so it can be differentiated using the\npower…

Explanation:

Step1: Polynomial long division

Divide (x^{2}-9x + 18) by (x - 4). [ \begin{align*} x^{2}-9x+18&=(x - 4)(x-5)-2\ \frac{x^{2}-9x + 18}{x - 4}&=\frac{(x - 4)(x - 5)-2}{x - 4}\ \end{align*} ]

Step2: Simplify the expression

[ \begin{align*} \frac{(x - 4)(x - 5)-2}{x - 4}&=\frac{(x - 4)(x - 5)}{x - 4}-\frac{2}{x - 4}\ &=x-5-\frac{2}{x - 4}\ &=x-5-2(x - 4)^{-1} \end{align*} ] Now, each term (x), (-5), (-2(x - 4)^{-1}) can be differentiated using the power rule. The power rule states that if (y = x^{n}), then (y^\prime=nx^{n - 1}), for (n\neq0). For (y=x) ((n = 1)), (y^\prime=1); for (y=-5) ((n = 0)), (y^\prime=0); for (y=-2(x - 4)^{-1}), using the chain - rule (a generalization of the power rule when we have a function inside another function, here (u=x - 4) and (y=-2u^{-1}), (y^\prime=\frac{dy}{du}\cdot\frac{du}{dx}), (\frac{dy}{du}=2u^{-2}), (\frac{du}{dx}=1), so (y^\prime = 2(x - 4)^{-2})).

For option A: ((x^{2}-9x + 18)(x - 4)^{-1}) would require the product rule (((uv)^\prime=u^\prime v+uv^\prime) where (u=x^{2}-9x + 18) and (v=(x - 4)^{-1})) to differentiate, not just the power rule.

For option C: (\frac{(x - 3)(x - 6)}{x - 4}=\frac{x^{2}-9x + 18}{x - 4}) is the original function and still in the quotient form, which would need either the quotient rule ((\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v-uv^\prime}{v^{2}}) where (u=x^{2}-9x + 18) and (v=x - 4)) or polynomial long - division (as we did for option B) to rewrite it for power - rule - based differentiation.

Answer:

B. (x - 5-\frac{2}{x - 4})