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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

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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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