math1a - quiz 7 - solns

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

This preview shows pages 1–2. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 09/17/2011 for the course MATH 1A taught by Professor Wilkening during the Spring '08 term at Berkeley.

### Page1 / 2

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online