-/1 points details my notes scalc9 2.9.033. use a linear approximation (or differentials) to estimate the…

Explanation:

Step1: Choose a function and a - value

Let $y = f(x)=\sqrt[3]{x}$, and choose $a = 216$ since $\sqrt[3]{216}=6$ is easy to calculate.

Step2: Find the derivative of the function

The derivative of $y = f(x)=x^{\frac{1}{3}}$ using the power - rule $(x^n)^\prime=nx^{n - 1}$ is $f^\prime(x)=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{x^{2}}}$.

Step3: Evaluate the derivative at $a$

$f^\prime(216)=\frac{1}{3\sqrt[3]{216^{2}}}=\frac{1}{3\times6^{2}}=\frac{1}{108}$.

Step4: Find $\Delta x$

$\Delta x=x - a$, where $x = 217$ and $a = 216$, so $\Delta x=217 - 216 = 1$.

Step5: Use the linear approximation formula

The linear approximation formula is $L(x)=f(a)+f^\prime(a)(x - a)$. Substituting $a = 216$, $x = 217$, $f(216)=6$, and $f^\prime(216)=\frac{1}{108}$ into the formula, we get $L(217)=f(216)+f^\prime(216)(217 - 216)=6+\frac{1}{108}\times1$.

Step6: Calculate the approximation

$L(217)=6+\frac{1}{108}\approx6 + 0.00926=6.00926$.

Answer:

$6.00926$