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\n b. determine an equation of the tangent line at p\n $f(x)=sqrt{5x + 39},p(5,8)$\n a. $m_{tan}square$\n (simplify your answer. type an exact answer, using radicals as needed)

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\n b. determine an equation of the tangent line at p\n $f(x)=sqrt{5x + 39},p(5,8)$\n a. $m_{tan}square$\n (simplify your answer. type an exact answer, using radicals as needed)

Answer

Explanation:

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

Given (f(x)=\sqrt{5x + 30}) and (a = 5), then (f(a+h)=f(5 + h)=\sqrt{5(5 + h)+30}=\sqrt{25+5h + 30}=\sqrt{5h + 55}), and (f(a)=f(5)=\sqrt{5\times5+30}=\sqrt{25 + 30}=\sqrt{55}=8).

Step2: Calculate the limit for the slope

[ \begin{align*} m_{\tan}&=\lim_{h\rightarrow0}\frac{f(a + h)-f(a)}{h}\ &=\lim_{h\rightarrow0}\frac{\sqrt{5h + 55}-\sqrt{55}}{h}\ &=\lim_{h\rightarrow0}\frac{(\sqrt{5h + 55}-\sqrt{55})(\sqrt{5h + 55}+\sqrt{55})}{h(\sqrt{5h + 55}+\sqrt{55})}\ &=\lim_{h\rightarrow0}\frac{(5h + 55)-55}{h(\sqrt{5h + 55}+\sqrt{55})}\ &=\lim_{h\rightarrow0}\frac{5h}{h(\sqrt{5h + 55}+\sqrt{55})}\ &=\lim_{h\rightarrow0}\frac{5}{\sqrt{5h + 55}+\sqrt{55}} \end{align*} ] Substitute (h = 0) into the limit: (\frac{5}{\sqrt{55}+\sqrt{55}}=\frac{5}{2\sqrt{55}})

Step3: 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)=(5,8)) and (m=\frac{5}{2\sqrt{55}}) [ \begin{align*} y-8&=\frac{5}{2\sqrt{55}}(x - 5)\ y&=\frac{5}{2\sqrt{55}}x-\frac{25}{2\sqrt{55}}+8 \end{align*} ]

Answer:

a. (\frac{5}{2\sqrt{55}}) b. (y=\frac{5}{2\sqrt{55}}x-\frac{25}{2\sqrt{55}}+8)