Question

question 3
let ( f(x)=-3x^{2}+18x + 7). which of the following is (\frac{f(x + h)-f(x)}{h})?

• ( - 6xh-3h^{2}+18h)
• ( - 6x^{2}+36x + 14+h)
• ( - 6x-3h + 18)
• ( - 3x^{2}-3h^{2}+18x + 18h+7)
• ( - 3x^{2}-6xh-3h^{2}+18x + 18h+7)

Explanation:

Step1: Find \(f(x + h)\)

Given \(f(x)=-3x^{2}+18x + 7\), then \(f(x + h)=-3(x + h)^{2}+18(x + h)+7\). Expand \((x + h)^{2}=x^{2}+2xh+h^{2}\), so \(f(x + h)=-3(x^{2}+2xh + h^{2})+18x+18h + 7=-3x^{2}-6xh-3h^{2}+18x + 18h+7\).

Step2: Calculate \(f(x + h)-f(x)\)

\(f(x + h)-f(x)=(-3x^{2}-6xh-3h^{2}+18x + 18h + 7)-(-3x^{2}+18x + 7)\). Remove the parentheses: \(-3x^{2}-6xh-3h^{2}+18x + 18h + 7 + 3x^{2}-18x - 7=-6xh-3h^{2}+18h\).

Step3: Calculate \(\frac{f(x + h)-f(x)}{h}\)

\(\frac{f(x + h)-f(x)}{h}=\frac{-6xh-3h^{2}+18h}{h}\). Factor out \(h\) from the numerator: \(\frac{h(-6x-3h + 18)}{h}\). Cancel out \(h\) (\(h
eq0\)), we get \(-6x-3h + 18\).

Answer:

\(-6x-3h + 18\)