Question

1. find the linear approximation for $y = sqrt3{x}$ for $a = 125$. then use it to estimate the following:
2. $sqrt3{126}$
3. $sqrt3{125.50}$

Explanation:

Step1: Recall linear - approximation formula

The linear - approximation formula is $L(x)=f(a)+f^{\prime}(a)(x - a)$, where $y = f(x)=\sqrt[3]{x}=x^{\frac{1}{3}}$, $a = 125$.

Step2: Find the derivative of $y=f(x)$

Using the power rule $(x^n)^\prime=nx^{n - 1}$, for $y=x^{\frac{1}{3}}$, $y^\prime=f^{\prime}(x)=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3\sqrt[3]{x^{2}}}$.

Step3: Evaluate $f(a)$ and $f^{\prime}(a)$

When $a = 125$, $f(125)=\sqrt[3]{125}=5$ and $f^{\prime}(125)=\frac{1}{3\sqrt[3]{125^{2}}}=\frac{1}{3\times25}=\frac{1}{75}$.

Step4: Find the linear - approximation $L(x)$

$L(x)=f(125)+f^{\prime}(125)(x - 125)=5+\frac{1}{75}(x - 125)=5+\frac{1}{75}x-\frac{5}{3}=\frac{1}{75}x+\frac{10}{3}$.

Step5: Estimate $\sqrt[3]{126}$

Let $x = 126$. Then $L(126)=\frac{1}{75}\times126+\frac{10}{3}=\frac{126}{75}+\frac{250}{75}=\frac{126 + 250}{75}=\frac{376}{75}\approx5.013$.

Step6: Estimate $\sqrt[3]{125.50}$

Let $x = 125.50$. Then $L(125.50)=\frac{1}{75}\times125.50+\frac{10}{3}=\frac{125.50}{75}+\frac{250}{75}=\frac{125.50+250}{75}=\frac{375.50}{75}\approx5.007$.

Answer:

For $\sqrt[3]{126}\approx5.013$, for $\sqrt[3]{125.50}\approx5.007$