Find the linearization of f, g, h at x=0. What do you notice about each expression? How do you explain what happened?

Question

Let 𝑓(𝑥) = (𝑥 − 1)² 𝑔(𝑥) = 𝑒^-2x ℎ(𝑥) = 1 + ln(1 − 2𝑥)

Answer 1

The linearization of a function at a given point, a, is:

L(x) = f(a) + f'(a)(x-a)

The procedure for each equation is to first evaluate f(x) and f'(x)

Evaluation each expression f(x), g(x), h(x) at x = 0 gives:

1. f(0) = (0 - 1)² = 1
2. g(0) = e^{-2 * 0} = e⁰ = 1
3. h(0) = 1 + ln(1 - 2(0) ) = 1 +ln(1) = 1 + 0 = 1

They're all equal! Now to derive and evaluate at 0

1. f'(x) = 2(x-1)(1)
2. f'(0) = 2(0 - 1) = -2
3. g'(x) = -2e^-2x
4. g'(0) = -2e^-2(0) = -2
5. h'(x) = ( 1/(1 - 2x) ) * -2
6. h'(0) = ( 1/(1 - 2(0) ) * -2 = (1/1) * -2 = -2

... And they're all equal again

So going back to our equation for L(x)

1. L_f(x) = 1 + -2(x - 0) = 1 - 2x
2. L_g(x) = 1 + -2(x - 0) = 1 - 2x
3. L_h(x) = 1 + -2(x - 0) = 1 - 2x

L(x) is equivalent for all 3 functions.

To explain this, plugging f(x), g(x), h(x), and L(x) into a graphing calculator can be helpful to visualize why.

A plot can be viewed here: https://www.desmos.com/calculator/jgnhtoqyne

While the 3 functions are divergent, they do converge at x=0 (L(x) is the green line)