webwork 3 - topics 6 - 7: problem 2 (1 point) use linear approximation to approximate √64.2 as follows. let…

Explanation:

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

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

Step2: Calculate the slope $m$

Evaluate $f'(x)$ at $x = 64$. $m=f'(64)=\frac{1}{2\sqrt{64}}=\frac{1}{2\times8}=\frac{1}{16}$.

Step3: Find the point on the function

When $x = 64$, $f(64)=\sqrt{64}=8$. So the point $(x_0,y_0)=(64,8)$ lies on the function $y = f(x)$.

Step4: Calculate the y - intercept $b$

Use the point - slope form $y - y_0=m(x - x_0)$ and rewrite it in $y=mx + b$ form. Substitute $x = 64$, $y = 8$, and $m=\frac{1}{16}$ into $y=mx + b$. We get $8=\frac{1}{16}\times64 + b$. Simplifying the right - hand side gives $8 = 4 + b$, so $b = 4$.

Step5: Approximate $\sqrt{64.2}$

Use the linear approximation $L(x)=m(x - x_0)+y_0$. Here $x = 64.2$, $x_0 = 64$, $m=\frac{1}{16}$, and $y_0 = 8$. So $L(64.2)=\frac{1}{16}(64.2 - 64)+8=\frac{1}{16}\times0.2+8=\frac{0.2}{16}+8=8+\frac{1}{80}=8.0125$.

Answer:

$m=\frac{1}{16}$ $b = 4$ Answer: $8.0125$