Question

homework4: problem 4 (1 point) let f = 2x^3 - 1 and find the values below 1. f(x + h) = 2. (f(x + h) - f(x)) = 3. lim_{h→0} (f(x + h) - f(x))/h = 4. find the equation of the line normal to the graph of f at x = 1. y =

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)$:
\[

$$\begin{align*} f(x + h)&=2(x + h)^{3}-1\\ &=2(x^{3}+3x^{2}h + 3xh^{2}+h^{3})-1\\ &=2x^{3}+6x^{2}h + 6xh^{2}+2h^{3}-1 \end{align*}$$

\]

Step2: Find $f(x + h)-f(x)$

\[

$$\begin{align*} f(x + h)-f(x)&=(2x^{3}+6x^{2}h + 6xh^{2}+2h^{3}-1)-(2x^{3}-1)\\ &=2x^{3}+6x^{2}h + 6xh^{2}+2h^{3}-1 - 2x^{3}+1\\ &=6x^{2}h+6xh^{2}+2h^{3} \end{align*}$$

\]

Step3: Find $\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}$
\[

$$\begin{align*} \lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h ightarrow0}\frac{6x^{2}h + 6xh^{2}+2h^{3}}{h}\\ &=\lim_{h ightarrow0}(6x^{2}+6xh + 2h^{2})\\ &=6x^{2} \end{align*}$$

\]

Step4: Find the equation of the normal - line

First, find the slope of the tangent line at $x = 1$.
When $x = 1$, the slope of the tangent line $m_{t}=6x^{2}\big|_{x = 1}=6$.
The slope of the normal line $m_{n}=-\frac{1}{6}$.
When $x = 1$, $f(1)=2\times1^{3}-1=1$.
Using the point - slope form $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(1,1)$ and $m =-\frac{1}{6}$, we get $y - 1=-\frac{1}{6}(x - 1)$.
\[y=-\frac{1}{6}x+\frac{1}{6}+1=-\frac{1}{6}x+\frac{7}{6}\]

Answer:

1. $2x^{3}+6x^{2}h + 6xh^{2}+2h^{3}-1$
2. $6x^{2}h+6xh^{2}+2h^{3}$
3. $6x^{2}$
4. $y =-\frac{1}{6}x+\frac{7}{6}$