Question

therefore, we have
m_{\tan}=lim_{x
ightarrow a}\frac{f(x)-f(a)}{x - a}
=lim_{x
ightarrow36}\frac{square-sqrt{36}}{x - 36}
=lim_{x
ightarrow36}\frac{square - 6}{x - 36}

Explanation:

Step1: Identify the function

Given the formula for the slope of the tangent line $m_{\tan}=\lim_{x
ightarrow a}\frac{f(x)-f(a)}{x - a}$, and $a = 36$, $f(a)=\sqrt{36}=6$. If we assume $f(x)=\sqrt{x}$, then we substitute $f(x)$ into the limit formula.

Step2: Fill in the blanks

The first blank should be $\sqrt{x}$ since $f(x)=\sqrt{x}$, and the second blank should also be $\sqrt{x}$ as we are using the function $f(x)=\sqrt{x}$ in the limit - based formula for the slope of the tangent line.

Answer:

The first blank: $\sqrt{x}$; The second blank: $\sqrt{x}$