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 $x = 0$.
$l(x)=\square$

Explanation:

Step1: Find the derivative of the function

The function is \(f(x)=x^{2}+4x\). Using the power rule \((x^n)^\prime = nx^{n - 1}\), the derivative \(f^\prime(x)=(x^{2}+4x)^\prime=2x + 4\).

Step2: Evaluate the function and its derivative at \(x = a=0\)

• Evaluate \(f(x)\) at \(x = 0\): \(f(0)=0^{2}+4\times0=0\).
• Evaluate \(f^\prime(x)\) at \(x = 0\): \(f^\prime(0)=2\times0 + 4=4\).

Step3: Use the linearization formula \(L(x)=f(a)+f^\prime(a)(x - a)\)

Since \(a = 0\), \(f(a)=0\) and \(f^\prime(a)=4\), substituting into the formula \(L(x)=f(a)+f^\prime(a)(x - a)\), we get \(L(x)=0+4(x - 0)\).

Answer:

\(L(x)=4x\)