find a linearization that will replace the function over an interval that includes the given point $x_0$…

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.\n$f(x)=x^{2}+4x,x_0 = 0.01$\nset the center of the linearization as $x = 0$.\n$l(x)=\\square$

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.\n$f(x)=x^{2}+4x,x_0 = 0.01$\nset the center of the linearization as $x = 0$.\n$l(x)=\\square$

Answer

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)