Question

find a linearization that will replace the function over an interval that includes the given point $x_0$. center the linearization not at $x_0$ but at a nearby integer, $x = a$, at which the given function and its derivative are easy to evaluate.

$f(x)=x^{2}+4x,x_0 = 0.01$

set the center of the linearization as $x=square$.

Explanation:

Step1: Analyze the given \(x_0\) value

Given \(x_0 = 0.01\). We need to find a nearby integer \(a\) such that \(f(a)\) and \(f^{\prime}(a)\) are easy to evaluate.

Step2: Determine the nearby integer

The integer closest to \(0.01\) is \(a = 0\).

• Calculate \(f(a)\):
• For \(f(x)=x^{2}+4x\) and \(a = 0\), \(f(0)=0^{2}+4\times0=0\).
• Calculate \(f^{\prime}(x)\) and \(f^{\prime}(a)\):
• Using the power rule \((x^{n})^{\prime}=nx^{n - 1}\), \(f^{\prime}(x)=\frac{d}{dx}(x^{2}+4x)=2x + 4\).
• When \(a = 0\), \(f^{\prime}(0)=2\times0+4 = 4\).

Answer:

\(0\)