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Unformatted text preview: MAT 319 Solutions for HW11 Exercise 1. Section 6.1, #16. Proof. Since the function tan is continuous and strictly increasing from ( π 2 , π 2 ) to R , we know that the inverse function arctan exists, and is continuous and strictly increasing from R to ( π 2 , π 2 ) . Moreover ( tan ) ′ ( x ) = ( cos x ) . ( cos x ) − ( sin x )( − sin x ) ( cos x ) 2 = 1 ( cos x ) 2 which is on ( π 2 , π 2 ) . Therefore we know that arctan is differentiable on R and that ( arctan x ) ′ = 1 ( tan ) ′ ( arctan x ) = 1 1 + ( tan ( arctan x )) 2 = 1 1 + x 2 , because 1 ( cos x ) 2 = sin 2 x + cos 2 x cos 2 x = tan 2 x + 1 square Exercise 2. Section 6.2, #1b. Proof. We have g ( x ) = 3 x 4 x 2 and g ′ ( x ) = 3 8 x which is < on (∞ , 3/8) , equal to 0 at x = 3/8 and > on (3/8 , + ∞ ) , therefore g is strictly decreasing before 3/8 and then strictly increasing after. Thus it has a single global minimum at 3/8 and this is also the only local extremum....
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This note was uploaded on 01/31/2011 for the course AMS 319 taught by Professor Mark during the Fall '10 term at SUNY Stony Brook.
 Fall '10
 Mark

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