# The linearization of the function h at the point x = -4 is L(x) = -(2x + 5). Let K be the...

## Question:

The linearization of the function {eq}\displaystyle h {/eq} at the point {eq}\displaystyle x = -4 {/eq} is {eq}\displaystyle h(x) = -(2x + 5) {/eq}. Let {eq}\displaystyle K {/eq} be the linearization of the function {eq}\displaystyle u(x) = xh(x) {/eq} at {eq}\displaystyle x = -4 {/eq}. Find {eq}\displaystyle K{/eq}.

## Linearization

Linear approximations are important to use if the function is very difficult to estimate on its own. One form that linear approximation or linearization is based off of is Euler's method, which is denoted as:

$$y_1 = y_0 + h \cdot y_0' $$

where {eq}y_0 {/eq}, {eq}h {/eq}, and {eq}y_0' {/eq}, have to be given or determined in order to the determine the unknown {eq}y_1 {/eq} value.

Linear approximation is different in that a {eq}y_0 {/eq} is needed to be discovered before figuring out the {eq}h {/eq} value and {eq}y_{0}' {/eq}. The linearization formula similarly is denoted as:

$$f(x) = f(x_0) + h \cdot f'(x_{0}) $$

where {eq}f(x)_0 {/eq}, {eq}h = x - x_0 {/eq}, and {eq}f'(x_{0}) {/eq}, have to be given or determined in order to the determine the unknown {eq}f(x) {/eq} value.

## Answer and Explanation:

1. List down all the given information.

$$\begin{align} u(x) &= xh(x) \\ &= x\cdot -(2x + 5) \\ \\ x &= -4 \end{align} $$

2. Solve for the general step size.

$$\begin{align} h &= x-(-4) \\ &= x+4 \end{align} $$

3. Solve the derivative of {eq}f(x) {/eq}.

$$\begin{align*} u'(x) = \frac{d}{dx}u(x) &= \frac{d}{dx}(x\cdot -(2x + 5)) \\ &= \frac{d}{dx}(-2x^2+5x) \\ &= 2\cdot -2x^{2-1}+5\cdot 1 \\ &= -4x+5 \end{align*} $$

4. Substitute the resulting values into the linearization formula to compute {eq}K(x) {/eq}.

$$\begin{align*} K(x) = u(-4)+h\cdot u'(-4) &= \left(-2(-4)^2+5(-4)\right)+(x+4)\cdot (-4(-4)+5) \\ &= \left(-2(16)-20 \right)+(x+4)\cdot (16+5) \\ &= (-32-20)+(x+4)\cdot (21) \\ &= -52+21x+84 \\ &= 21x+32 \\ \end{align*} $$

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