Finding an Equation of a Tangent Line
- Find the derivative of \(f( x) =\sqrt{2x}\) at \(x=8.\)
- Use the derivative \(f^\prime ( 8)\) to find the equation of the tangent line to the graph of \(f\) at the point (8,4).
Solution (a) The derivative of \(f\) at \(8\) is \[ \begin{eqnarray*} f^\prime (8) &=& \lim\limits_{x\rightarrow 8}\dfrac{f( x) -f( 8) }{x-8} {=} \lim\limits_{x\rightarrow 8}\dfrac{ \sqrt{2x}-4}{x-8} {=} \lim\limits_{x\rightarrow 8}\dfrac{( \sqrt{2x}-4) ( \sqrt{2x}+4) }{( x-8) ( \sqrt{2x}+4) }\\[-6.4pt] &&\hspace{4.0pc}\underset{\color{#0066A7}{f(8) =\sqrt{2\cdot 8} = 4}}{\color{#0066A7}{\uparrow}}\hspace{1.1pc} \underset{\underset{\color{#0066A7}{\scriptsize \hspace{0.4pc}\hbox{the numerator.}}}{\color{#0066A7}{\scriptsize \hbox{Rationalize}}}}{\color{#0066A7}{\uparrow}}\\ &=& \lim\limits_{x\rightarrow 8}\dfrac{2x-16}{( x-8) ( \sqrt{ 2x}+4) } =\lim\limits_{x\rightarrow 8}\dfrac{2( x-8) }{ ( x-8) ( \sqrt{2x}+4) }=\lim\limits_{x\rightarrow 8} \dfrac{2}{\sqrt{2x}+4}=\dfrac{1}{4} \end{eqnarray*} \]
(b) The slope of the tangent line to the graph of \(f\) at the point \((8,4) \) is \(f^\prime (8) =\dfrac{1}{4}.\) Using the point-slope form of a line, we get \[ \begin{eqnarray*} \begin{array}{rl@{\qquad}l} y-f( 8) &= f^\prime ( 8) ( x-8) & {\color{#0066A7}{\hbox{\({y-y}_{1}=m ( {x-x}_{1})\)}}} \\[6pt] \notag y-4 &= \dfrac{1}{4}( x-8) & {\color{#0066A7}{\hbox{\({f}( {8}) = 4; \quad {f^\prime }(8) = \dfrac{1}{4}\)}}} \\[9pt] \notag y &= \dfrac{1}{4}x-\dfrac{1}{4}\cdot 8+4 \\[9pt] \notag y &= \dfrac{1}{4}x+2 \end{array} \end{eqnarray*} \]
