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Find the linearization of \(\displaystyle f(x) = \sqrt{x}\) at \(\displaystyle x = 9\) and use it to approximate \(\displaystyle \sqrt{6}\). Leave your solution as a rational number.

The formula for the linearization of \(\displaystyle f\) at \(\displaystyle x=a\) is: \(\displaystyle L(x)=f'(a)(x-a)+f(a)\). \(\displaystyle f(9) = 3\) and caclulating the derivative gives \(\displaystyle f'(x) = \frac{1}{2 \sqrt{x}}\) and \(\displaystyle f'(9)=\frac{1}{6}\). The linearization is \(\displaystyle L(x)=\frac{1}{6} \left(x - 9\right) + 3 = \frac{x}{6} + \frac{3}{2}\)This gives the approximation \(\displaystyle \sqrt{6} \approx L(6) = \frac{5}{2}\).

Download \(\LaTeX\) 

\begin{question}Find the linearization of $f(x) = \sqrt{x}$ at $x = 9$ and use it to approximate $\sqrt{6}$. Leave your solution as a rational number. 
    \soln{9cm}{The formula for the linearization of $f$ at $x=a$ is: $L(x)=f'(a)(x-a)+f(a)$. $f(9) = 3$ and caclulating the derivative gives $f'(x) = \frac{1}{2 \sqrt{x}}$ and $f'(9)=\frac{1}{6}$. The linearization is $L(x)=\frac{1}{6} \left(x - 9\right) + 3 = \frac{x}{6} + \frac{3}{2}$This gives the approximation $\sqrt{6} \approx L(6) = \frac{5}{2}$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)
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HTML for Canvas 
<p> <p>Find the linearization of  <img class="equation_image" title=" \displaystyle f(x) = \sqrt{x} " src="/equation_images/%20%5Cdisplaystyle%20f%28x%29%20%3D%20%5Csqrt%7Bx%7D%20" alt="LaTeX:  \displaystyle f(x) = \sqrt{x} " data-equation-content=" \displaystyle f(x) = \sqrt{x} " />  at  <img class="equation_image" title=" \displaystyle x = 9 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%209%20" alt="LaTeX:  \displaystyle x = 9 " data-equation-content=" \displaystyle x = 9 " />  and use it to approximate  <img class="equation_image" title=" \displaystyle \sqrt{6} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B6%7D%20" alt="LaTeX:  \displaystyle \sqrt{6} " data-equation-content=" \displaystyle \sqrt{6} " /> . Leave your solution as a rational number. </p> </p>
HTML for Canvas
<p> <p>The formula for the linearization of  <img class="equation_image" title=" \displaystyle f " src="/equation_images/%20%5Cdisplaystyle%20f%20" alt="LaTeX:  \displaystyle f " data-equation-content=" \displaystyle f " />  at  <img class="equation_image" title=" \displaystyle x=a " src="/equation_images/%20%5Cdisplaystyle%20x%3Da%20" alt="LaTeX:  \displaystyle x=a " data-equation-content=" \displaystyle x=a " />  is:  <img class="equation_image" title=" \displaystyle L(x)=f'(a)(x-a)+f(a) " src="/equation_images/%20%5Cdisplaystyle%20L%28x%29%3Df%27%28a%29%28x-a%29%2Bf%28a%29%20" alt="LaTeX:  \displaystyle L(x)=f'(a)(x-a)+f(a) " data-equation-content=" \displaystyle L(x)=f'(a)(x-a)+f(a) " /> .  <img class="equation_image" title=" \displaystyle f(9) = 3 " src="/equation_images/%20%5Cdisplaystyle%20f%289%29%20%3D%203%20" alt="LaTeX:  \displaystyle f(9) = 3 " data-equation-content=" \displaystyle f(9) = 3 " />  and caclulating the derivative gives  <img class="equation_image" title=" \displaystyle f'(x) = \frac{1}{2 \sqrt{x}} " src="/equation_images/%20%5Cdisplaystyle%20f%27%28x%29%20%3D%20%5Cfrac%7B1%7D%7B2%20%5Csqrt%7Bx%7D%7D%20" alt="LaTeX:  \displaystyle f'(x) = \frac{1}{2 \sqrt{x}} " data-equation-content=" \displaystyle f'(x) = \frac{1}{2 \sqrt{x}} " />  and  <img class="equation_image" title=" \displaystyle f'(9)=\frac{1}{6} " src="/equation_images/%20%5Cdisplaystyle%20f%27%289%29%3D%5Cfrac%7B1%7D%7B6%7D%20" alt="LaTeX:  \displaystyle f'(9)=\frac{1}{6} " data-equation-content=" \displaystyle f'(9)=\frac{1}{6} " /> . The linearization is  <img class="equation_image" title=" \displaystyle L(x)=\frac{1}{6} \left(x - 9\right) + 3 = \frac{x}{6} + \frac{3}{2} " src="/equation_images/%20%5Cdisplaystyle%20L%28x%29%3D%5Cfrac%7B1%7D%7B6%7D%20%5Cleft%28x%20-%209%5Cright%29%20%2B%203%20%3D%20%5Cfrac%7Bx%7D%7B6%7D%20%2B%20%5Cfrac%7B3%7D%7B2%7D%20" alt="LaTeX:  \displaystyle L(x)=\frac{1}{6} \left(x - 9\right) + 3 = \frac{x}{6} + \frac{3}{2} " data-equation-content=" \displaystyle L(x)=\frac{1}{6} \left(x - 9\right) + 3 = \frac{x}{6} + \frac{3}{2} " /> This gives the approximation  <img class="equation_image" title=" \displaystyle \sqrt{6} \approx L(6) = \frac{5}{2} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B6%7D%20%5Capprox%20L%286%29%20%3D%20%5Cfrac%7B5%7D%7B2%7D%20" alt="LaTeX:  \displaystyle \sqrt{6} \approx L(6) = \frac{5}{2} " data-equation-content=" \displaystyle \sqrt{6} \approx L(6) = \frac{5}{2} " /> . </p> </p>