math1a - quiz 7 - solns

math1a - quiz 7 - solns - -2 x 1/x Solution First write(1-2...

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Math 1A Quiz 7 Solutions March 21st, 2008 1. Let f ( x ) = tanh - 1 (sin x ). a) What is the domain of f ? b) Find a formula for f 0 ( x ). Simplify as much as possible. Solution to a): For x to be in the domain of f ( x ) = tanh - 1 (sin x ), we must have that: (a) x is in the domain of sin, and (b) sin x is in the domain of tanh - 1 . Since the domain of sin is R , we just need sin x to be in the domain of tanh - 1 , which is ( - 1 , 1). So we want - 1 < sin x < 1. Since the range of sin is [ - 1 , 1], we just want sin x 6 = ± 1. So the domain is x 6 = π 2 + . We could also write that the domain is ··· ∪ ± - 3 π 2 , - π 2 ² ³ - π 2 , π 2 ´ ± π 2 , 3 π 2 ² ∪ ··· Solution to b): We use the identity d dx tanh - 1 ( x ) = 1 1 - x 2 , plus chain rule: f 0 ( x ) = 1 1 - sin 2 ( x ) cos x. Since 1 - sin 2 ( x ) = cos 2 x , f 0 ( x ) = 1 cos 2 ( x ) cos x = 1 cos x . 2. Find lim x 0 (1
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Unformatted text preview: -2 x ) 1 /x . Solution: First write (1-2 x ) 1 /x = e (1 /x ) ln(1-2 x ) . We want to figure out what happens to this exponent as x goes to 0, i.e. we want to find: lim x → ln(1-2 x ) x . 1 Now, as x → 0, we have ln(1-2 x ) → ln 1 = 0 and x → 0, so we can apply L’Hospital’s rule to this limit, and we get: lim x → ln(1-2 x ) x = lim x →-2 1-2 x 1 = lim x →-2 1-2 x =-2 . So the exponent in our original expression e (1 /x ) ln(1-2 x ) goes to-2 as x goes to 0. Hence lim x → (1-2 x ) 1 /x = lim x → e (1 /x ) ln(1-2 x ) = e-2 . 2...
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This note was uploaded on 09/17/2011 for the course MATH 1A taught by Professor Wilkening during the Spring '08 term at Berkeley.

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math1a - quiz 7 - solns - -2 x 1/x Solution First write(1-2...

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