let f(x)=1/(3 + 2x) and point p=(1,1/5). a. use the following definition of the slope of the tangent line at…

let f(x)=1/(3 + 2x) and point p=(1,1/5). a. use the following definition of the slope of the tangent line at x = a to find the slope of the line tangent to the graph of f at p. m_tan=lim_(h→0) (f(a + h)-f(a))/h b. determine an equation of the tangent line at p. a. the slope of the tangent line is □. (simplify your answer.)

let f(x)=1/(3 + 2x) and point p=(1,1/5). a. use the following definition of the slope of the tangent line at x = a to find the slope of the line tangent to the graph of f at p. m_tan=lim_(h→0) (f(a + h)-f(a))/h b. determine an equation of the tangent line at p. a. the slope of the tangent line is □. (simplify your answer.)

Answer

Answer:

a. $-\frac{2}{25}$ b. $y =-\frac{2}{25}x+\frac{7}{25}$

Explanation:

Step1: Find $f(a + h)$ and $f(a)$

Given $f(x)=\frac{1}{3 + 2x}$, $a = 1$. Then $f(1+h)=\frac{1}{3+2(1 + h)}=\frac{1}{5 + 2h}$ and $f(1)=\frac{1}{5}$.

Step2: Substitute into slope - formula

$m_{\tan}=\lim_{h\rightarrow0}\frac{f(1 + h)-f(1)}{h}=\lim_{h\rightarrow0}\frac{\frac{1}{5 + 2h}-\frac{1}{5}}{h}$. First, find a common denominator: $\frac{1}{5 + 2h}-\frac{1}{5}=\frac{5-(5 + 2h)}{5(5 + 2h)}=\frac{-2h}{5(5 + 2h)}$. So, $m_{\tan}=\lim_{h\rightarrow0}\frac{\frac{-2h}{5(5 + 2h)}}{h}=\lim_{h\rightarrow0}\frac{-2h}{5h(5 + 2h)}$.

Step3: Simplify the limit

Cancel out the $h$ terms: $m_{\tan}=\lim_{h\rightarrow0}\frac{-2}{5(5 + 2h)}$. Substitute $h = 0$: $m_{\tan}=-\frac{2}{25}$.

Step4: Find the equation of the tangent line

Use the point - slope form $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,\frac{1}{5})$ and $m =-\frac{2}{25}$. $y-\frac{1}{5}=-\frac{2}{25}(x - 1)$. Expand: $y-\frac{1}{5}=-\frac{2}{25}x+\frac{2}{25}$. Add $\frac{1}{5}=\frac{5}{25}$ to both sides: $y=-\frac{2}{25}x+\frac{2 + 5}{25}=-\frac{2}{25}x+\frac{7}{25}$.