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Math191_Week6_section

# Math191_Week6_section - -π 2 π 2 and the range to-∞ ∞...

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Math 191 Problem Set for Week 6 Section 1. Which of the following functions have inverses? a. f ( x ) = ( x - 1) 3 with domain ( -∞ , ). f ( x ) is invertible because it is a strictly increasing function (that is, f ( a ) > f ( b ) whenever a > b ). b. f ( x ) = 1 2+sin x with domain [ - π/ 2 ,π/ 2]. f ( x ) is invertible as it is one-to-one: if f ( a ) = f ( b ) then 1 2+sin a = 1 2+sin b and so sin a = sin b and so a = b , as sin x is one-to-one on the domain [ - π/ 2 ,π/ 2]. c. f ( x ) = sec 2 x with domain ( - π/ 2 ,π/ 2). f ( x ) is not invertible as it is not one-to-one: f ( x ) = f ( - x ) for all x 6 = 0. 2. a. Show that the function f ( x ) = x 3 + 3 x is one-to-one and hence has an inverse f - 1 ( x ) . b. What are the domain and the range of f - 1 ( x ) ? c. Find d dx f - 1 ( x ) at x = 4 . 3. Let f ( x ) = sin - 1 (tan x ) . a. What is the natural domain for f ( x ) so that the inverse function f - 1 ( x ) exists? Step 1. The function f ( x ) must be one-to-one so that the inverse function f - 1 ( x ) exists. The function f ( x ) is one-to-one if and only if the functions sin - 1 x and tan x are one-to-one. Step 2. The domain D 1 of tan x is ( -∞ , ). The range R 1 is ( -∞ , ). However we must restrict the function to be one-to-one so that the inverse will exist. Thus, we set the domain to
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Unformatted text preview: (-π 2 , π 2 ) and the range to (-∞ , ∞ ). Similarly, the domain D 2 of sin-1 x is [-1 , 1]. The range R 2 is [-π 2 , π 2 ]. The function sin-1 x is one-to-one in the domain D 2 and the range R 2 , and so the inverse exists. Step 3. The range R 1 of tan x must agree with the domain D 2 of sin-1 x , i.e., R 1 = D 2 = [-1 , 1]. To achieve this we will restrict the domain D 1 of tan x to [-π 4 , π 4 ]. Thus, the domain of f ( x ) is [-π 4 , π 4 ] and the range is [-π 2 , π 2 ]. b. Calculate the inverse function f-1 ( x ) and specify its domain. The inverse function f-1 ( x ) = tan-1 (sin x ). Pulling directly from the range above, the domain is [-π 2 , π 2 ]. 4. Evaluate R dx 5 x √ ln(3 x ) . 5. Evaluate ln x x (1+ln x ) dx . 6. Evaluate R √ ln π 2 xe x 2 cos( e x 2 ) dx . 1...
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