Question

1. -/1 points find the linearization l(x) of the function at a. f(x) = \\(\sqrt3{x}\\), a = 64 l(x) =

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 $f(x)=\sqrt[3]{x}=x^{\frac{1}{3}}$ and $a = 64$. Then $f(a)=f(64)=\sqrt[3]{64}=4$.

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

Using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, for $f(x)=x^{\frac{1}{3}}$, we have $f^{\prime}(x)=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3x^{\frac{2}{3}}}$.

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

Substitute $x = 64$ into $f^{\prime}(x)$. So $f^{\prime}(64)=\frac{1}{3\times64^{\frac{2}{3}}}=\frac{1}{3\times16}=\frac{1}{48}$.

Step5: Calculate $L(x)$

Substitute $f(a) = 4$, $f^{\prime}(a)=\frac{1}{48}$ and $a = 64$ into the linearization formula $L(x)=f(a)+f^{\prime}(a)(x - a)$.
$L(x)=4+\frac{1}{48}(x - 64)=4+\frac{1}{48}x-\frac{4}{3}=\frac{1}{48}x+\frac{8}{3}$.

Answer:

$L(x)=\frac{1}{48}x+\frac{8}{3}$