Question

quiz 6 - requires respondus lockdown browser + webcam
started: sep 27 at 12:25pm
quiz instructions
access code: start
timed: 30 minutes
number of attempts: 1
lockdown browser and respondus monitor required.
this quiz covers material from this week (sections 2.7, 2.8, 2.9).
question 4
find the linearization, l(x), of the function at a.
f(x) = √x where a = 9
o x/6 + 3/2
o x/18 + 2/3
o x/2 + 2/3
o the correct answer is not listed.

Explanation:

Step1: Recall linearization formula

The linearization $L(x)$ of a function $y = f(x)$ at $x = a$ is given by $L(x)=f(a)+f^{\prime}(a)(x - a)$.

Step2: Find $f(a)$

Given $f(x)=\sqrt{x}=x^{\frac{1}{2}}$ and $a = 9$. Then $f(a)=f(9)=\sqrt{9}=3$.

Step3: Find the derivative of $f(x)$

Using the power - rule $(x^n)^\prime=nx^{n - 1}$, for $f(x)=x^{\frac{1}{2}}$, we have $f^{\prime}(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$.

Step4: Find $f^{\prime}(a)$

Substitute $x = a = 9$ into $f^{\prime}(x)$. So $f^{\prime}(9)=\frac{1}{2\sqrt{9}}=\frac{1}{6}$.

Step5: Calculate $L(x)$

$L(x)=f(9)+f^{\prime}(9)(x - 9)$. Substitute $f(9) = 3$ and $f^{\prime}(9)=\frac{1}{6}$ into the formula:
\[

$$\begin{align*} L(x)&=3+\frac{1}{6}(x - 9)\\ &=3+\frac{1}{6}x-\frac{9}{6}\\ &=\frac{1}{6}x+\frac{18 - 9}{6}\\ &=\frac{1}{6}x+\frac{3}{2} \end{align*}$$

\]

Answer:

A. $\frac{x}{6}+\frac{3}{2}$