Exam3_2 2 - Taylor Douglas Exam 3 Due Dec 5 2007 1:00 am...

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Taylor, Douglas – Exam 3 – Due: Dec 5 2007, 1:00 am – Inst: JEGilbert 2 7 x θ p x 2 + 7 From this it follows that y = sin θ = x x 2 + 7 . Alternatively, we can use the trig identity csc 2 θ = 1 + cot 2 θ to determine sin θ . keywords: 004 (part 1 of 1) 10 points Find the inverse function, f - 1 , of f when f is de±ned by f ( x ) = 6 x - 5 , x 5 6 . 1. f - 1 ( x ) = 1 5 p x 2 + 6 , x 6 5 2. f - 1 ( x ) = 1 6 p x 2 - 5 , x 0 3. f - 1 ( x ) = 1 6 ( x 2 + 5) , x 0 correct 4. f - 1 ( x ) = 1 5 p x 2 - 6 , x 0 5. f - 1 ( x ) = 1 5 ( x 2 - 6) , x 5 6 6. f - 1 ( x ) = 1 6 ( x 2 + 5) , x 6 5 Explanation: Since f has domain [ 5 6 , ) and is increasing on its domain, the inverse of f exists and has range [ 5 6 , ); furthermore, since f has range [0 , ), the inverse of f has domain [0 , ). To determine f - 1 we solve for x in y = 6 x - 5 and then interchange x, y
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