a. use the definition $m_{tan}=lim_{h \to 0}\frac{f(a + h)-f(a)}{h}$ to find the slope of the line tangent…

a. use the definition $m_{tan}=lim_{h \to 0}\frac{f(a + h)-f(a)}{h}$ to find the slope of the line tangent to the graph of f at p\nb. determine an equation of the tangent line at p\n$f(x)=\frac{1}{2 + 3x},p(1,\frac{1}{5})$\noa $lim_{h \to 0}\frac{1}{1 + h}$\nob $lim_{h \to 0}\frac{f(1 + h)-\frac{1}{5}}{h}$\noc $lim_{h \to 0}\frac{f(1 + h)+\frac{1}{5}}{h}$\nod $lim_{h \to 0}\frac{\frac{1}{5}-f(1 + h)}{h}$\nthe slope of the tangent line is (simplify your answer.)

a. use the definition $m_{tan}=lim_{h \to 0}\frac{f(a + h)-f(a)}{h}$ to find the slope of the line tangent to the graph of f at p\nb. determine an equation of the tangent line at p\n$f(x)=\frac{1}{2 + 3x},p(1,\frac{1}{5})$\noa $lim_{h \to 0}\frac{1}{1 + h}$\nob $lim_{h \to 0}\frac{f(1 + h)-\frac{1}{5}}{h}$\noc $lim_{h \to 0}\frac{f(1 + h)+\frac{1}{5}}{h}$\nod $lim_{h \to 0}\frac{\frac{1}{5}-f(1 + h)}{h}$\nthe slope of the tangent line is (simplify your answer.)

Answer

Explanation:

Step1: Recall the slope - of - tangent formula

The formula for the slope of the tangent line to the graph of $y = f(x)$ at the point $(a,f(a))$ is $m_{tan}=\lim_{h\rightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $a = 1$ and $f(x)=\frac{1}{2 + 3x}$, so $f(a)=f(1)=\frac{1}{2+3\times1}=\frac{1}{5}$. The correct limit expression for the slope of the tangent line at the point $P(1,\frac{1}{5})$ is $\lim_{h\rightarrow0}\frac{f(1 + h)-\frac{1}{5}}{h}$, which is option B.

Step2: Find $f(1 + h)$

$f(1 + h)=\frac{1}{2+3(1 + h)}=\frac{1}{2 + 3+3h}=\frac{1}{5 + 3h}$.

Step3: Calculate the limit for the slope

$m_{tan}=\lim_{h\rightarrow0}\frac{f(1 + h)-\frac{1}{5}}{h}=\lim_{h\rightarrow0}\frac{\frac{1}{5 + 3h}-\frac{1}{5}}{h}$. First, find a common denominator: $\frac{1}{5 + 3h}-\frac{1}{5}=\frac{5-(5 + 3h)}{5(5 + 3h)}=\frac{5 - 5-3h}{5(5 + 3h)}=\frac{-3h}{5(5 + 3h)}$. So, $\lim_{h\rightarrow0}\frac{\frac{1}{5 + 3h}-\frac{1}{5}}{h}=\lim_{h\rightarrow0}\frac{\frac{-3h}{5(5 + 3h)}}{h}=\lim_{h\rightarrow0}\frac{-3h}{5h(5 + 3h)}$. Cancel out the $h$ terms: $\lim_{h\rightarrow0}\frac{-3}{5(5 + 3h)}=-\frac{3}{25}$.

Step4: Find the equation of the tangent line

The point - slope form of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(1,\frac{1}{5})$ and $m =-\frac{3}{25}$. $y-\frac{1}{5}=-\frac{3}{25}(x - 1)$. Multiply through by 25 to clear the fractions: $25y-5=-3(x - 1)$. Expand: $25y-5=-3x + 3$. Rearrange to the slope - intercept form $y=mx + b$: $25y=-3x+8$, so $y=-\frac{3}{25}x+\frac{8}{25}$.

Answer:

The correct option for the limit expression of the slope is B. The slope of the tangent line is $-\frac{3}{25}$. The equation of the tangent line is $y =-\frac{3}{25}x+\frac{8}{25}$