Question

math 104
question 3. let f(x)=2x² + 3x. if h≠0, then the difference quotient can be simplified as
\frac{f(x + h)-f(x)}{h}=ah + bx + c,
where a, b, and c are constants. (note: it is possible for one or more of these constants to be 0.)
(a) simplify the difference quotient to find the constants a, b, and c.
(b) use the simplified expression, found in part (a), to find f(x)=lim_{h \to 0}\frac{f(x + h)-f(x)}{h}.
(c) find each of the following values.
i. f(1) ii. f(2) iii. f(3)

Explanation:

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

Given \(f(x)=2x^{2}+3x\), then \(f(x + h)=2(x + h)^{2}+3(x + h)=2(x^{2}+2xh+h^{2})+3x + 3h=2x^{2}+4xh+2h^{2}+3x + 3h\)

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

\[

$$\begin{align*} f(x + h)-f(x)&=(2x^{2}+4xh+2h^{2}+3x + 3h)-(2x^{2}+3x)\\ &=2x^{2}+4xh+2h^{2}+3x + 3h - 2x^{2}-3x\\ &=4xh+2h^{2}+3h \end{align*}$$

\]

Step3: Simplify the difference - quotient \(\frac{f(x + h)-f(x)}{h}\)

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{4xh+2h^{2}+3h}{h}\\ &=4x + 2h+3 \end{align*}$$

\]
Comparing with \(Ah + Bx + C\), we have \(A = 2\), \(B = 4\), \(C = 3\)

Step4: Find \(f^{\prime}(x)\)

\[

$$\begin{align*} f^{\prime}(x)&=\lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}\\ &=\lim_{h ightarrow0}(4x + 2h+3)\\ &=4x+3 \end{align*}$$

\]

Step5: Find \(f^{\prime}(1)\), \(f^{\prime}(2)\) and \(f^{\prime}(3)\)

• For \(x = 1\), \(f^{\prime}(1)=4\times1+3=7\)
• For \(x = 2\), \(f^{\prime}(2)=4\times2+3=11\)
• For \(x = 3\), \(f^{\prime}(3)=4\times3+3=15\)

Answer:

(a) \(A = 2\), \(B = 4\), \(C = 3\)
(b) \(f^{\prime}(x)=4x + 3\)
(c)
i. \(f^{\prime}(1)=7\)
ii. \(f^{\prime}(2)=11\)
iii. \(f^{\prime}(3)=15\)