Question

use linear approximation to approximate √16.4 as follows. let f(x)=√x. the equation of the tangent line to f(x) at x = 16 can be written in the form y=mx + b. compute m and b. m = b = using this find the approximation for √16.4. answer.

Explanation:

Step1: Find the derivative of $f(x)$

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

Step2: Calculate the slope $m$ at $x = 16$

Substitute $x = 16$ into $f^\prime(x)$. Then $m=f^\prime(16)=\frac{1}{2\sqrt{16}}=\frac{1}{2\times4}=\frac{1}{8}$.

Step3: Calculate $f(16)$

$f(16)=\sqrt{16}=4$.

Step4: Find the $y -$intercept $b$

The equation of the tangent line is $y=mx + b$. Substitute $x = 16$, $y = 4$ and $m=\frac{1}{8}$ into it. We get $4=\frac{1}{8}\times16 + b$. Solving for $b$: $4 = 2 + b$, so $b = 2$.

Step5: Use linear approximation

The linear approximation of $y = f(x)$ near $x=a$ is $L(x)=f(a)+f^\prime(a)(x - a)$. Here $a = 16$ and $x=16.4$. The linear approximation formula $y=mx + b$ with $m=\frac{1}{8}$ and $b = 2$ gives $y=\frac{1}{8}x+2$. Substitute $x = 16.4$ into it: $y=\frac{1}{8}\times16.4+2=\frac{16.4}{8}+2=2.05 + 2=4.05$.

Answer:

$m=\frac{1}{8}$, $b = 2$, and the approximation for $\sqrt{16.4}$ is $4.05$