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Find the linearization of \(\displaystyle f(x) = \sqrt{x}\) at \(\displaystyle x = 4\) and use it to approximate \(\displaystyle \sqrt{3}\). Leave your solution as a rational number.

The formula for the linearization of \(\displaystyle f\) at \(\displaystyle x=a\) is: \(\displaystyle L(x)=f'(a)(x-a)+f(a)\). \(\displaystyle f(4) = 2\) and caclulating the derivative gives \(\displaystyle f'(x) = \frac{1}{2 \sqrt{x}}\) and \(\displaystyle f'(4)=\frac{1}{4}\). The linearization is \(\displaystyle L(x)=\frac{1}{4} \left(x - 4\right) + 2 = \frac{x}{4} + 1\)This gives the approximation \(\displaystyle \sqrt{3} \approx L(3) = \frac{7}{4}\).

Download \(\LaTeX\) 

\begin{question}Find the linearization of $f(x) = \sqrt{x}$ at $x = 4$ and use it to approximate $\sqrt{3}$. Leave your solution as a rational number. 
    \soln{9cm}{The formula for the linearization of $f$ at $x=a$ is: $L(x)=f'(a)(x-a)+f(a)$. $f(4) = 2$ and caclulating the derivative gives $f'(x) = \frac{1}{2 \sqrt{x}}$ and $f'(4)=\frac{1}{4}$. The linearization is $L(x)=\frac{1}{4} \left(x - 4\right) + 2 = \frac{x}{4} + 1$This gives the approximation $\sqrt{3} \approx L(3) = \frac{7}{4}$. }

\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 = 4 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%204%20" alt="LaTeX:  \displaystyle x = 4 " data-equation-content=" \displaystyle x = 4 " />  and use it to approximate  <img class="equation_image" title=" \displaystyle \sqrt{3} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B3%7D%20" alt="LaTeX:  \displaystyle \sqrt{3} " data-equation-content=" \displaystyle \sqrt{3} " /> . 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(4) = 2 " src="/equation_images/%20%5Cdisplaystyle%20f%284%29%20%3D%202%20" alt="LaTeX:  \displaystyle f(4) = 2 " data-equation-content=" \displaystyle f(4) = 2 " />  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'(4)=\frac{1}{4} " src="/equation_images/%20%5Cdisplaystyle%20f%27%284%29%3D%5Cfrac%7B1%7D%7B4%7D%20" alt="LaTeX:  \displaystyle f'(4)=\frac{1}{4} " data-equation-content=" \displaystyle f'(4)=\frac{1}{4} " /> . The linearization is  <img class="equation_image" title=" \displaystyle L(x)=\frac{1}{4} \left(x - 4\right) + 2 = \frac{x}{4} + 1 " src="/equation_images/%20%5Cdisplaystyle%20L%28x%29%3D%5Cfrac%7B1%7D%7B4%7D%20%5Cleft%28x%20-%204%5Cright%29%20%2B%202%20%3D%20%5Cfrac%7Bx%7D%7B4%7D%20%2B%201%20" alt="LaTeX:  \displaystyle L(x)=\frac{1}{4} \left(x - 4\right) + 2 = \frac{x}{4} + 1 " data-equation-content=" \displaystyle L(x)=\frac{1}{4} \left(x - 4\right) + 2 = \frac{x}{4} + 1 " /> This gives the approximation  <img class="equation_image" title=" \displaystyle \sqrt{3} \approx L(3) = \frac{7}{4} " src="/equation_images/%20%5Cdisplaystyle%20%5Csqrt%7B3%7D%20%5Capprox%20L%283%29%20%3D%20%5Cfrac%7B7%7D%7B4%7D%20" alt="LaTeX:  \displaystyle \sqrt{3} \approx L(3) = \frac{7}{4} " data-equation-content=" \displaystyle \sqrt{3} \approx L(3) = \frac{7}{4} " /> . </p> </p>