find the linearization l(x) of the function at a. f(x) = ∛x, a = 64 l(x) = (√x / 3)(x - 64) × need help…

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 $(x^n)^\prime=nx^{n - 1}$, for $f(x)=x^{\frac{1}{3}}$, we have $f^{\prime}(x)=\frac{1}{3}x^{-\frac{2}{3}}$.

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

Substitute $x = 64$ into $f^{\prime}(x)$. So $f^{\prime}(64)=\frac{1}{3}(64)^{-\frac{2}{3}}=\frac{1}{3}(4^{3})^{-\frac{2}{3}}=\frac{1}{3}\times4^{-2}=\frac{1}{3}\times\frac{1}{16}=\frac{1}{48}$.

Step5: Calculate $L(x)$

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

Answer:

$L(x)=4+\frac{1}{48}(x - 64)$