use linear approximation to estimate ( 2.8^{3} ) as follows.\nlet ( f(x)=x^{3} ). the equation of the…

use linear approximation to estimate ( 2.8^{3} ) as follows.\nlet ( 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\n\n( m=quad ) and ( b=quad ).\n\nthus, an approximation for ( 2.8^{3} ) is\n\n

use linear approximation to estimate ( 2.8^{3} ) as follows.\nlet ( 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\n\n( m=quad ) and ( b=quad ).\n\nthus, an approximation for ( 2.8^{3} ) is\n\n

Answer

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

[ \begin{align*} y-27&=27x-81\ y&=27x- 54 \end{align*} ] 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). [ \begin{align*} y&=75.6-54\ y&=21.6 \end{align*} ]

Answer:

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