## Precalculus (6th Edition) Blitzer

a) The slope of the tangent to the graph of $f\left( x \right)=\sqrt{x}$ at $\left( 16,4 \right)$ is $\frac{1}{8}$. b) The slope intercept equation of the tangent line is $y=\frac{1}{8}x+2$.
(a) Consider the function $f\left( x \right)=\sqrt{x}$ and the point $\left( 16,4 \right)$ Here, $\left( a,f\left( a \right) \right)=\left( 16,4 \right)$ Now, compute the slope of the tangent line for the function $f\left( x \right)=\sqrt{x}$ as follows: Put $a=16$ , ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f\left( 16+h \right)-f\left( 16 \right)}{h}$ To compute $f\left( 16+h \right)$, substitute $x=16+h$ in the function $f\left( x \right)=\sqrt{x}$. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{16+h}-\sqrt{16}}{h}$ Now, multiply and divide with the conjugate of $\sqrt{16+h}-\sqrt{16}$ that is $\sqrt{16+h}+\sqrt{16}$ and simplify using the property $\left( A-B \right)\left( A+B \right)={{A}^{2}}-{{B}^{2}}$. \begin{align} & {{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{\sqrt{16+h}-\sqrt{16}}{h}\cdot \frac{\sqrt{16+h}+\sqrt{16}}{\sqrt{16+h}+\sqrt{16}} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{{{\left( \sqrt{16+h} \right)}^{2}}-16}{h\left( \sqrt{16+h}+4 \right)} \\ & =\underset{h\to 0}{\mathop{\lim }}\,\frac{16+h-16}{h\left( \sqrt{16+h}+4 \right)} \end{align} Combine the like terms in the numerator and cancel out h. ${{m}_{\tan }}=\underset{h\to 0}{\mathop{\lim }}\,\frac{1}{\left( \sqrt{16+h}+4 \right)}$ Apply the limits and simplify. \begin{align} & {{m}_{\tan }}=\frac{1}{\left( \sqrt{16+0}+4 \right)} \\ & =\frac{1}{4+4} \\ & =\frac{1}{8} \end{align} Thus, the slope of the tangent to the graph of $f\left( x \right)=\sqrt{x}$ at $\left( 16,4 \right)$ is $\frac{1}{8}$. (b) Consider the function $f\left( x \right)=\sqrt{x}$ and the point $\left( 16,4 \right)$ From part (a), the slope of the tangent to the graph of $f\left( x \right)=\sqrt{x}$ at $\left( 16,4 \right)$ is $\frac{1}{8}$. Now compute the slope intercept equation of the tangent line as follows: Here, the tangent line passes through the point $\left( 16,4 \right)$ which implies ${{x}_{1}}=16$ and ${{y}_{1}}=4$ Substitute $m=\frac{1}{8},{{x}_{1}}=16,{{y}_{1}}=4$ in the equation $y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$. \begin{align} & y-4=\frac{1}{8}\left( x-16 \right) \\ & y-4=\frac{1}{8}x-2 \\ & y=\frac{1}{8}x-2+4 \\ & y=\frac{1}{8}x+2 \end{align}