Question

In Exercises \(1-4,\) find a first-degree polynomial function \(P_{1}\) whose value and slope agree with the value and slope of \(f\) at \(x=c .\) Use a graphing utility to graph \(f\) and \(P_{1} .\) What is \(P_{1}\) called? $$ f(x)=\frac{4}{\sqrt[3]{x}}, \quad c=8 $$

Short answer

The first-degree polynomial function \(P_{1}\) that has the same value and slope as function \(f\) at \(x=8\) is \(P_{1}(x) = -\frac{1}{12}x + \frac{8}{3}\). Also known as the tangent line at \(x=c\), in this case at \(x=8\).

Step-by-step solution

Determine the value of \(f\) at \(x=c\)

First, substitute \(x=c\) into \(f\) to get its value at \(c\). It will be \(f(c) = \frac{4}{\sqrt[3]{8}} = 2\).

Determine the slope of \(f\) at \(x=c\)

We first need to find \(f'\), the derivative of \(f(x)\). Using the rule for differentiating \(x^{-n}\), we get \(f'(x) = - \frac{4} {3x^{\frac{4}{3}}}\). Now substitute \(x=c\) into the derivative to get the slope at point \(c\) , which will be: \(f'(c) = -\frac{4}{3*\sqrt[3]{8^4}} = -\frac{1}{12}\).

Find the first-degree polynomial function \(P_{1}\)

A first degree polynomial function is of the form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept. We want this polynomial to have the same value and slope as function \(f\) at \(x=c\), therefore we substitute these values into the equation to solve for \(b\). This gives: \(2 = -\frac{1}{12} * 8 + b\). Solving for \(b\) we get \(b = 2 + \frac{2}{3} = \frac{8}{3}\), Hence the polynomial is \(P_{1}(x) = -\frac{1}{12}x + \frac{8}{3}\).