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))…

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 $\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$

[ \begin{align*} \lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h\rightarrow0}\frac{6x^{2}h + 6xh^{2}+2h^{3}}{h}\ &=\lim_{h\rightarrow0}(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. $y =-\frac{1}{6}x+\frac{7}{6}$