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Let $ f(x) = (x - 1)^2 g(x) = e^{-2x} $and $ h(x) = 1 + \ln(1 -2x) $(a) Find the linearization of $ f, g $ and $ h $ at $ a = 0. $ What do you notice? How do you explain what happened?(b) Graph $ f, g, $ and $ h $ and their linear approximations. For which function is the linear approximation best? For which is it worst? Explain.

(a) SEE EXPLANATION(b) If $x>0,$ then $f, g, h$ going from best to worst. If $x<0,$ then $h, f, g$ goingfrom best to worst. Please see graph

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 10

Linear Approximation and Differentials

Derivatives

Differentiation

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Okay. We know that because each part of X is native to over one plus two acts, you know h of zero is one, and hpe from zero is gonna be negative to. Therefore, we have one month's two acts. Recall the old three liberalizations are equivalent to each other because G prime of zero is the same as all the other variables. Promise here, As you can see, if X is greater than zero than F G and H go from best to worst. If X is less than zero after an edge or at h f G go from best to worst.

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