Question

use linear approximation, i.e. the tangent line, to approximate \\( \sqrt 3 { 64.2 } \\) as follows: let \\( f ( x ) = \sqrt 3 { x } \\). the equation of the tangent line to \\( f ( x ) \\) at \\( x = 64 \\) can be written in the form \\( y = m x + b \\) where \\( m \\) is: and where \\( b \\) is: using this, we find our approximation for \\( \sqrt 3 { 64.2 } \\) is question help: video message instructor submit question jump to answer

Explanation:

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

The function \(f(x)=\sqrt[3]{x}=x^{\frac{1}{3}}\).
Using the power rule \((x^n)^\prime = nx^{n - 1}\), we have \(f^\prime(x)=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3x^{\frac{2}{3}}}\).
When \(x = 64\), \(m=f^\prime(64)=\frac{1}{3\times64^{\frac{2}{3}}}\).
Since \(64^{\frac{2}{3}}=(64^{\frac{1}{3}})^2 = 4^2=16\), then \(m=\frac{1}{3\times16}=\frac{1}{48}\).

Step2: Find the value of \(f(64)\)

\(f(64)=\sqrt[3]{64}=4\).
Using the point - slope form \(y - y_0=m(x - x_0)\) (here \(x_0 = 64,y_0 = 4\)), \(y-4=\frac{1}{48}(x - 64)\).
Rewrite it in \(y=mx + b\) form: \(y=\frac{1}{48}x+4-\frac{64}{48}\).
Simplify \(4-\frac{64}{48}=\frac{192 - 64}{48}=\frac{128}{48}=\frac{8}{3}\), so \(b=\frac{8}{3}\).

Step3: Approximate \(\sqrt[3]{64.2}\)

Let \(x = 64.2\), then \(y=\frac{1}{48}\times64.2+\frac{8}{3}\).
\(\frac{1}{48}\times64.2=\frac{64.2}{48}=\frac{642}{480}=\frac{107}{80}=1.3375\), \(\frac{8}{3}\approx2.6667\).
\(y=1.3375 + 2.6667=4.0042\)

Answer:

\(m=\frac{1}{48}\), \(b = \frac{8}{3}\), \(\sqrt[3]{64.2}\approx4.0042\)