Download Calculus I by Paul Dawkins PDF

By Paul Dawkins

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Solution First, before we even start solving we need to make one thing clear. DO NOT CANCEL AN x FROM BOTH SIDES!!! While this may seem like a natural thing to do it WILL cause us to lose a solution here. So, to solve this equation we’ll first get all the terms on one side of the equation and then factor an x out of the equation. If we can cancel an x from all terms then it can be factored out. Doing this gives, 5 x tan ( 8 x ) - 3x = x ( 5 tan ( 8 x ) - 3) = 0 Upon factoring we can see that we must have either, x=0 tan ( 8 x ) = OR 3 5 Note that if we’d canceled the x we would have missed the first solution.

4p 2p ( 2 ) 16p + = < 2p 15 5 15 p 2p ( 2 ) 17p x= + = < 2p 3 5 15 x= n = 3. 4p 2p ( 3) 22p + = < 2p 15 5 15 p 2p ( 3) 23p x= + = < 2p 3 5 15 x= n = 4. 4p 2p ( 4 ) 28p + = < 2p 15 5 15 p 2p ( 4 ) 29p = < 2p x= + 3 5 15 x= n = 5. 4p 2p ( 5 ) 34p + = > 2p 15 5 15 p 2p ( 5 ) 35p x= + = > 2p 3 5 15 x= Okay, so we finally got past the right endpoint of our interval so we don’t need any more positive n. Now let’s take a look at the negative n and see what we’ve got. n = –1 . aspx Calculus I n = –2. 4p 2p ( -2 ) 8p + => -p 15 5 15 p 2p ( -2 ) 7p => -p x= + 3 5 15 x= n = –3.

X f(x) g(x) -2 1 f ( -2 ) = 2 = 4 æ1ö g ( -2 ) = ç ÷ = 4 è2ø -1 f ( -1) = 2 -1 = 1 2 æ1ö g ( -1) = ç ÷ = 2 è2ø -2 0 f ( 0 ) = 20 = 1 1 f (1) = 2 2 f ( 2) = 4 -2 -1 0 æ1ö g ( 0) = ç ÷ = 1 è2ø 1 g (1) = 2 1 g ( 2) = 4 Here’s the sketch of both of these functions. aspx Calculus I This graph illustrates some very nice properties about exponential functions in general. Properties of f ( x ) = b x 1. f ( 0 ) = 1 . The function will always take the value of 1 at x = 0 . 2. f ( x ) ¹ 0 . An exponential function will never be zero.

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