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

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

Answer

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:

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