MAC
c1-t2-a(1)

# Consequently applying theorem 255 with f as above g x

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Consequently, applying Theorem 2.5.5 with f as above, g (x) = sin -1 ( x ) and c = 0, since sin -1 ( x ) 0 when x 0, we have lim x 0 x sin 1 ( x ) lim x 0 sin(sin 1 ( x )) sin 1 ( x ) lim x 0 f ( g ( x )) f (0) 1.

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TEST2/MAC2311 Page 3 of 4 ______________________________________________________________________ 5. (10 pts.) What are the x- and y - intercepts of the tangent line to the graph of y = 1/x 2 at the point (2,1/4)? The slope of the tangent line at x = 2 is dy dx x 2 2 x 3 x 2 1 4 . Consequently, with a little routine algebra, we see that an equation for the tangent line in slope-intercept form is given by y 1 4 x 3 4 . The y intercept is plainly y = 3/4 and one can see with a little work that the x intercept is x = 3.// ______________________________________________________________________ 6. (10 pts.) (a) Find all values in the interval [0,2 π ] at which the graph of f has a horizontal tangent line when f ( x ) = x + 2 cos( x ). Since f has a horizontal tangent line when f ( x ) = 0, and f ( x ) = 1 - 2sin( x ), it follows that f has a horizontal tangent line in the interval [0,2 π ] when sin( x ) = 1/2. This happens only at x 1 = π /6 or x 2 = 5 π /6 . (b) The following limit represents f ( a ) for some function f and some number a . Using that information, evaluate the limit. lim x → π sin(3 x ) 0 x π 3. Here, of course, f ( x ) = sin(3 x ) and a = π . So we have lim x → π sin(3 x ) 0 x π f ( π ) 3cos(3 π ) 3( 1) 3 since f ( x ) 3cos(3 x ). ______________________________________________________________________ 7. (10 pts.) (a) Using complete sentences and appropriate notation, provide the precise mathematical definition for the derivative, f ( x ), of a function f ( x ).// The function f defined by the equation f ( x ) lim h 0 f ( x h ) f ( x ) h .
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