find a linearization that will replace the function over an interval that includes the given point x₀…

find a linearization that will replace the function over an interval that includes the given point x₀. center the linearization not at x₀ but at a nearby integer, x = a, at which the given function and its derivative are easy to evaluate. f(x)=x² + 5x, x₀ = 0.01. set the center of the linearization as x = 0. l(x) =

find a linearization that will replace the function over an interval that includes the given point x₀. center the linearization not at x₀ but at a nearby integer, x = a, at which the given function and its derivative are easy to evaluate. f(x)=x² + 5x, x₀ = 0.01. set the center of the linearization as x = 0. l(x) =

Answer

Explanation:

Step1: Find the derivative of the function

The function is $f(x)=x^{2}+5x$. Using the power - rule, if $y = x^n$, then $y^\prime=nx^{n - 1}$. So $f^\prime(x)=2x + 5$.

Step2: Evaluate the function and its derivative at the center point

The center point is $a = 0$. Evaluate $f(0)$ and $f^\prime(0)$. For $f(x)=x^{2}+5x$, when $x = 0$, $f(0)=0^{2}+5\times0=0$. For $f^\prime(x)=2x + 5$, when $x = 0$, $f^\prime(0)=2\times0+5 = 5$.

Step3: Use the linearization formula

The linearization formula is $L(x)=f(a)+f^\prime(a)(x - a)$. Substituting $a = 0$, $f(0)=0$, and $f^\prime(0)=5$ into the formula, we get $L(x)=0+5(x - 0)=5x$.

Answer:

$5x$