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? $$ f(x)=\frac{4}{\sqrt{x}}, \quad c=1 $$

Short answer

The first-degree polynomial function \(P_1(x)\) is \(P_1(x) = 4 - 2 * (x - 1) = 6 - 2x\). This function is called the tangent line or linear approximation to \(f(x)\) at \(x=c\).

Step-by-step solution

Find the Function Value at a Given Point

Substitute \(x=c\) into the function \(f (x) = \frac{4}{\sqrt{x}}\) to find \(f(c)\). This gives \(f(1) = \frac{4}{\sqrt{1}} = 4\).

Find the Function's Derivative

To find the slope of the function at the point \(x=c\), derive the function. The derivative of \(f(x) = \frac{4}{\sqrt{x}}\) is \(f'(x) =\frac{-2}{x^{3/2}}\).

Find the Derivative Value at the Given Point

Substitute \(x=c\) into the derivative \(f'(x) =\frac{-2}{x^{3/2}}\) to find \(f'(c)\). This gives \(f'(1) = \frac{-2}{1^{3/2}} = -2\).

Define the First-Degree Polynomial

The first-degree polynomial function \(P_1\) that matches \(f\) in both value and slope at \(x=c\) is \(P_1 (x) = f(c) + f'(c) * (x-c)\). Substitute \(c=1, f(c) = 4\) and \(f'(c) = -2\) into the formula to find \(P_1 (x)\). This gives \(P_1(x) = 4 - 2 * (x - 1)\).