Question

use linear approximation to estimate ( 2.8^{3} ) as follows.
let ( f(x)=x^{3} ). the equation of the tangent line to ( f(x) ) at ( x = 3 ) can be written in the form ( y=m x+b ) where

( m=quad ) and ( b=quad ).

thus, an approximation for ( 2.8^{3} ) is

Explanation:

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

The derivative of \(f(x)=x^{3}\) is \(f^{\prime}(x) = 3x^{2}\).

Step2: Calculate the slope \(m\)

When \(x = 3\), \(m=f^{\prime}(3)=3\times3^{2}=27\).

Step3: Find the value of \(f(3)\)

\(f(3)=3^{3}=27\).

Step4: Use the point - slope form \(y - y_{0}=m(x - x_{0})\)

Here \(x_{0}=3,y_{0}=27,m = 27\). The point - slope form is \(y-27=27(x - 3)\).

Step5: Rewrite in \(y=mx + b\) form

$$
LATEXBLOCK0
$$

So \(b=-54\).

Step6: Use linear approximation

We want to estimate \(2.8^{3}\). Let \(x = 2.8\). Using \(y=27x-54\), when \(x = 2.8\), \(y=27\times2.8-54\).

$$
LATEXBLOCK1
$$

Answer:

\(m = 27\), \(b=-54\), an approximation for \(2.8^{3}\) is \(21.6\)