Let f x x 3 x 2 2 x on 11 observe that f 1 2 f 1 0

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Let f ( x ) = x 3 + x 2 - 2 x on [-1,1]. Observe that f (-1) = 2, f (1) = 0, and k = 1 is a number between f (-1) and f (1). Since f is a polynomial, f is continuous on the interval [-1,1]. Consequently, the function f satisfies the hypotheses of the Intermediate Value Theorem. Thus, we are entitled to invoke the magical conclusion that asserts that there is at least one number x 0 in (-1,1) where f( x 0 ) = 1. ______________________________________________________________________ 3. (5 pts.) Pretend f is a magical function that has the property that at x = 3 the tangent line f is actually defined by the equation y = -2(x - 1) + 5. Obtain (a) f (3) = -2(3 - 1) + 5 = 1 (b) f (3) = -2
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TEST2/MAC2311 Page 2 of 4 ______________________________________________________________________ 4. (25 pts.) Compute the first derivatives of the following functions. You may use any of the rules of differentiation that are at your disposal. Do not attempt to simplify the algebra in your answers. (a) f ( x ) 4 x 6 7 x 12 8tan( x ) f ( x ) 24 x 5 84 x 13 8sec 2 ( x ) (b) g ( x ) (4 x 2 2 x 1 )sec( x ) g ( x ) (8 x 2 x 2 )sec( x ) (4 x 2 2 x 1 )sec( x )tan( x ) (c) h ( t ) 5 t 10 1 sin( t ) 2 h ( t ) (50 t 9 )(sin( t ) 2) (5 t 10 1)cos( t ) (sin( t ) 2) 2 (d) y cot 5 (2 θ 1) dy d θ 5(cot(2 θ 1)) 4 ( csc 2 (2 θ 1))(2) 10cot 4 (2 θ 1) csc 2 (2 θ 1) (e) L ( z ) sin(4 z 8 ) 4 csc ( π 6 ) 4cos( z 2 ) dL dz ( z ) cos(4 z 8 )(32 z 7 ) 0 4( sin( z 2 ))( 1 2 ) 32 z 7 cos(4 z 8 ) 2sin( z 2 ) ______________________________________________________________________ Silly 10 point Bonus Problem: (Solution) Theorem 2.5.5 asserts that if g is a function with and f is a function that is continuous at L , then lim x c g ( x ) L lim x c f ( g ( x )) f ( L ). It follows from that the function f defined by lim x 0 sin( x ) x 1. is continuous at x = 0. Since sine is continuous and one-to-one on [- π /2, π /2], sin -1 is one-to-one and continuous on [-1,1], and thus, lim x 0 sin 1 ( x ) sin 1 (0) 0.
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