Question

instructions
no credit without correct, thorough, mathematical work.
note: correct answers without correct supporting work receive no credit.

• work this problem out on 1 - sheet of paper, one - side only. label the problem, \defn of derivative\
• you will make a pdf of your all of your work - outs after you complete the test;
• make this problem page 1.
• after you submit this test, then submit your pdf in the canvas portal titled, \exam 1 work - out questions\
• an instructor will grade your work - out questions.

work - out problem 1, 15 points
a. (10 points) use the definition of the derivative ($f(x)=lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$) to compute $f$ of the function $f(x)=7 + 3x-2x^{2}$. no credit whatsoever for derivative obtained without using the definition.
b. (5 points) write the equation of the line tangent to the function $f(x)=7 + 3x-2x^{2}$ at $x = 1$. express the equation in slope - intercept form: $y=mx + b$.

Explanation:

Step1: Find $f(x + h)$

Given $f(x)=7 + 3x-2x^{2}$, then $f(x + h)=7+3(x + h)-2(x + h)^{2}=7 + 3x+3h-2(x^{2}+2xh+h^{2})=7 + 3x+3h-2x^{2}-4xh - 2h^{2}$.

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

$f(x + h)-f(x)=(7 + 3x+3h-2x^{2}-4xh - 2h^{2})-(7 + 3x-2x^{2})=3h-4xh - 2h^{2}$.

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

$\frac{f(x + h)-f(x)}{h}=\frac{3h-4xh - 2h^{2}}{h}=3-4x - 2h$.

Step4: Find the derivative $f'(x)$

$f'(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}(3-4x - 2h)=3-4x$.

Step5: Find the slope of the tangent - line at $x = 1$

When $x = 1$, $f'(1)=3-4\times1=-1$.

Step6: Find the value of $y$ at $x = 1$

$f(1)=7+3\times1-2\times1^{2}=7 + 3-2=8$.

Step7: Find the $y$-intercept $b$

Using the point - slope form $y - y_{1}=m(x - x_{1})$ with $(x_{1},y_{1})=(1,8)$ and $m=-1$, we have $y - 8=-1(x - 1)$.
Expanding to slope - intercept form: $y=-x + 9$.

Answer:

A. $f'(x)=3-4x$
B. $y=-x + 9$