Question

this weeks group project is only partially with the group. the following problem you can do together. (there will be one on the test like it!) 1. use the limit definition of the derivative to find the derivative function for f(x)=3x² - 2x - 10. then find an equation of the line tangent to the graph of f at x = 1. to earn all the points for this group project, the following needs to be completed individually. 2. go to the taft college library and pick a learning calculus book (not a textbook). there are some ebooks, general circulation books, and reserve books available. (**reserve books may be checked out for 2 hours, but may not be removed from the library.) some titles in the library are listed below. you may also find one of your own, but check with me first to make sure it is acceptable. read the section(s) that introduce derivative. write 2 or 3 paragraphs that summarize how they explained the derivative and your analysis of whether it helped your understanding of the concept of a derivative or not. include whether you think this would be a reference that you might use in the future to help your understanding of later topics in calculus. calculus for dummies the humongous book of calculus problems: translated for people who dont speak math calculus workbook for dummies homework helpers. (calculus qa300.s9852007) the manga guide to calculus (regular and ebook) the calculus lifesaver: all the tools you need to excel at calculus (qa 303.2.b362007)

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=3x^{2}-2x - 10$, we first find $f(x + h)$:
\[

$$\begin{align*} f(x + h)&=3(x + h)^{2}-2(x + h)-10\\ &=3(x^{2}+2xh+h^{2})-2x-2h - 10\\ &=3x^{2}+6xh+3h^{2}-2x-2h - 10 \end{align*}$$

\]

Step2: Substitute into the derivative formula

\[

$$\begin{align*} f^\prime(x)&=\lim_{h ightarrow0}\frac{(3x^{2}+6xh+3h^{2}-2x-2h - 10)-(3x^{2}-2x - 10)}{h}\\ &=\lim_{h ightarrow0}\frac{3x^{2}+6xh+3h^{2}-2x-2h - 10 - 3x^{2}+2x + 10}{h}\\ &=\lim_{h ightarrow0}\frac{6xh+3h^{2}-2h}{h}\\ &=\lim_{h ightarrow0}(6x + 3h-2) \end{align*}$$

\]
As $h
ightarrow0$, $f^\prime(x)=6x-2$.

Step3: Find the slope of the tangent line at $x = 1$

Substitute $x = 1$ into $f^\prime(x)$. Then $f^\prime(1)=6\times1-2=4$.

Step4: Find the point on the graph of $f(x)$ at $x = 1$

Substitute $x = 1$ into $f(x)$. $f(1)=3\times1^{2}-2\times1-10=3 - 2-10=-9$. The point is $(1,-9)$.

Step5: Use the point - slope form to find the tangent line equation

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(1,-9)$ and $m = 4$.
\[

$$\begin{align*} y-(-9)&=4(x - 1)\\ y + 9&=4x-4\\ y&=4x-13 \end{align*}$$

\]

Answer:

The derivative function is $f^\prime(x)=6x - 2$, and the equation of the tangent line to the graph of $f$ at $x = 1$ is $y=4x-13$.