Let f(x) = (x − 1)2, g(x) = e−2x, and h(x) = 1 + ln(1 − 2x)

Question

(a) Find the linearizations of f, g, and h at a = 0.

 What do you notice? How do you explain what happened? Lf (x) = Lg (x) = Lh (x) =

Answer 1

f(0) = g(0) = h(0) = 1

The respective derivatives at 0 all equal -2.

For example:

f'(x) = (x - 1)

g'(x) = -2 e^-2x

h'(x) = -2/(1-2x)

So the linearization of all three at a = 0 is the same for all three functions.

L(x) = -2(x-0) + 1, where -2 is the slope of the tangent line at the point (0,1) and 1 is the y-coordinate of the y-intercept. Using point-slope results in the same: y - 1 = -2(x-0) => y = -2x+1.

Check it out here:

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