Question

use linear approximation, i.e. the tangent line, to approximate \\(\sqrt3{1.3}\\) as follows:\
let \\(f(x) = \sqrt3{x}\\). the equation of the tangent line to \\(f(x)\\) at \\(x = 1\\) can be written in the form\
y = mx + b\
\
where m is:\
and where b is:\
\
using this, we find our approximation for \\(\sqrt3{1.3}\\) is\
\
question help: video message instructor

Explanation:

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

The function is \( f(x) = \sqrt[3]{x}=x^{\frac{1}{3}} \). The derivative \( f'(x) \) using the power rule \( \frac{d}{dx}x^n = nx^{n - 1} \) is \( f'(x)=\frac{1}{3}x^{-\frac{2}{3}}=\frac{1}{3x^{\frac{2}{3}}} \).

Step2: Calculate the slope \( m \) at \( x = 1 \)

Substitute \( x = 1 \) into \( f'(x) \). So \( m=f'(1)=\frac{1}{3\times1^{\frac{2}{3}}}=\frac{1}{3} \).

Step3: Find the value of \( f(1) \)

Substitute \( x = 1 \) into \( f(x) \), \( f(1)=\sqrt[3]{1}=1 \).

Step4: Calculate the y - intercept \( b \)

Using the point - slope form of a line \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(1,f(1))=(1,1) \) and \( m=\frac{1}{3} \). The equation of the line is \( y - 1=\frac{1}{3}(x - 1) \). Simplify it to slope - intercept form \( y=mx + b \): \( y=\frac{1}{3}x-\frac{1}{3}+1=\frac{1}{3}x+\frac{2}{3} \). So \( b = \frac{2}{3} \).

Step5: Approximate \( \sqrt[3]{1.3} \)

Use the tangent line equation \( y=\frac{1}{3}x+\frac{2}{3} \). Substitute \( x = 1.3 \) into the equation: \( y=\frac{1}{3}(1.3)+\frac{2}{3}=\frac{1.3 + 2}{3}=\frac{3.3}{3}=1.1 \).

Answer:

The value of \( m \) is \( \frac{1}{3} \), the value of \( b \) is \( \frac{2}{3} \), and the approximation for \( \sqrt[3]{1.3} \) is \( 1.1 \)