# Let f(x) = (x - 1)^2, g(x) = e^{- 2x}, and h(x) = 1 + \ln(1 - 2x). a) Find the linearizations...

## Question:

Let {eq}f(x) = (x - 1)^2, g(x) = e^{- 2x}, \ \mathrm{and} \ h(x) = 1 + \ln(1 - 2x). {/eq}

a) Find the linearizations of {eq}f, g, {/eq} and {eq}h {/eq} at {eq}a = 0. {/eq}

## Linearization:

Given that functions are often times difficult to evaluate as currently expressed, linearization allows us to estimate the value of the function. Mathmatically, the linearization formula is shown as {eq}f(x) = f(x_0) + h \cdot f'(x_{0}) {/eq}, 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:

Given: {eq}f(x) = (x-1)^2, g(x) = e^{-2x}, h(x) = 1+\ln(1-2x), a = 0 {/eq}

For {eq}f(x) = (x-1)^2 {/eq}:

The point of reference will be at {eq}x = 0 {/eq}:

The step size, accordingly, will be {eq}h = x-0 \Rightarrow h = x {/eq}

The derivative of {eq}f(x) {/eq} in general is

{eq}\begin{align*} f'(x) &= \frac{d}{dx}(f(x)) \\ &= \frac{d}{dx}((x-1)^2) \\ &= 2\cdot (x-1)^{2-1}\cdot 1 \\ &= 2(x-1) \\ &= 2x-2 \\ \end{align*} {/eq}

Now all the components to compute {eq}f(x) {/eq} are found, so simply apply the linear approximation formula to find {eq}L(x) {/eq}.

{eq}\begin{align*} L(x) = f(0)+h\cdot f'(0) &= (0-1)^2+x(2(0)-2) \\ &= (-1)^2+x(0-2) \\ &= 1+x(-2) \\ &= 1-2x \\ \end{align*} {/eq}

For {eq}g(x) = e^{-2x} {/eq}:

The point of reference will be at {eq}x = 0 {/eq}:

The step size, accordingly, will be {eq}h = x-0 \Rightarrow h = x {/eq}

The derivative of {eq}g(x) {/eq} in general is

{eq}\begin{align*} g'(x) &= \frac{d}{dx}(g(x)) \\ &= \frac{d}{dx}(e^{-2x}) \\ &= e^{-2x}\cdot -2 \\ &= -2e^{-2x} \\ \end{align*} {/eq}

Now all the components to compute {eq}g(x) {/eq} are found, so simply apply the linear approximation formula to find {eq}L(x) {/eq}.

{eq}\begin{align*} L(x) = g(0)+h\cdot g'(0) &= e^{-2(0)}+x(-2e^{-2(0)}) \\ &= e^{0}+x(-2e^{0}) \\ &= 1+x(-2(1)) \\ &= 1+x(-2) \\ &= 1-2x \\ \end{align*} {/eq}

For {eq}h(x) = 1+\ln(1-2x) {/eq}:

The point of reference will be at {eq}x = 0 {/eq}:

The step size, accordingly, will be {eq}h = x-0 \Rightarrow h = x {/eq}

The derivative of {eq}h(x) {/eq} in general is

{eq}\begin{align*} h'(x) &= \frac{d}{dx}(h(x)) \\ &= \frac{d}{dx}(1+\ln(1-2x)) \\ &= 0+\frac{1}{1-2x}\cdot -2 \\ &= -\frac{2}{1-2x} \\ \end{align*} {/eq}

Now all the components to compute {eq}h(x) {/eq} are found, so simply apply the linear approximation formula to find {eq}L(x) {/eq}.

{eq}\begin{align*} L(x) = h(0)+h\cdot h'(0) &= (1+\ln(1-2(0)))+x(-\frac{2}{1-2(0)}) \\ &= (1+\ln(1-0))+x(-\frac{2}{1-0}) \\ &= (1+\ln(1))+x(-\frac{2}{1}) \\ &= (1+0)+x(-2) \\ &= 1-2x \\ \end{align*} {/eq}