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ass8solns - Math 137 Assignment No 8 Solutions Fall 2011 1...

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Math 137 Assignment No. 8 - Solutions Fall 2011 1. Required to prove the identity arcsin parenleftbigg x 1 x + 1 parenrightbigg = 2 arctan x π 2 . (1) Because of the appearance of the square root, we consider only non-negative values of x , i.e., x 0. Define the function f ( x ) = arcsin parenleftbigg x 1 x + 1 parenrightbigg 2 arctan x + π 2 , x 0 . If we can show that f ( x ) = 0 for all x 0 then the identity in (1) follows. Note that f (0) = arcsin( 1) 2 arctan(0) + π 2 = π 2 + 0 + π 2 = 0 . We now compute f ( x ) by means of the Chain Rule, using the following derivatives, d du arcsin u = 1 1 u 2 , d du arctan u = 1 1 + u 2 . We find, after a little algebra, that d dx arcsin x 1 x + 1 = x + 1 2 x · 2 ( x + 1) 2 = 1 x ( x + 1) d dx arctan x = 1 1 + x · 1 2 x = 1 2 x ( x + 1) . This implies that f ( x ) = 0 for all x > 0. We can now employ Theorem 8 of Stewart’s text (which is based on the Mean Value Theorem), Section 4.2, Page 287, to conclude that f ( x ) = f (0) = 0 for all x 0, thus proving the identity in (1). For those who prefer to prove the desired result without recourse to Theorem 8 of the textbook, we may proceed as follows: For any x > 0, define the closed interval I = [0 , x ]. The function f is continuous on [0 , x ] and we have shown that f ( t ) = 0 for all t (0 , x ). As such, the hypotheses of the Mean Value Theorem are satisfied and we may conclude that f ( x ) f (0) x 0 = f ( x ) x = f ( c ) for some c (0 , x ) . But f ( c ) = 0 which implies that f ( x ) = 0. 1
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2. This question is concerned with the function f ( x ) = 3 x 2 / 3 x . (a) Required to find intercepts, intervals where f ( x ) > 0 and intervals where f ( x ) < 0. It helps if we examine the first term in the definition of f ( x ), which we’ll call g ( x ) = 3 x 2 / 3 . Note that we may express g ( x ) as 3( x 1 / 3 ) 2 . The graph of the function x 1 / 3 may be obtained from the graph of x 3 by reflecting the latter about the line y = x . By “squaring” the resulting graph (which essentially means reflecting any negative values about the x -axis to make them positive) and multiplying by 3 we obtain the graph of g ( x ) which is expected to have a cusp at x = 0. The graph is shown below. -5 -4 -3 -2 -1 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 The above analysis also shows that f ( x ) is defined for all x R . Now rewrite f ( x ) as f ( x ) = x 2 / 3 (3 x 1 / 3 ) , which implies that f ( x ) = 0 when x = 0 or x = 27. And since x 2 / 3 > 0 for all x negationslash = 0, it follows that f ( x ) > 0 when x 1 / 3 < 3. The latter inequality is satisfied for all x < 27 (the cube root of a negative number is a negative number) so if we exclude the point x = 0, we have the result that f ( x ) > 0 over the two intervals (i) ( −∞ , 0) and (ii) (0 , 27). (b) Regarding asymptotes, note that we may express f ( x ) as follows, f ( x ) = x (3 x 1 / 3 1) . As x → ±∞ , the term 3 x 1 / 3 0, implying that 3 x 1 / 3 1 ≈ − 1 for | x | sufficiently large. As such, we may conclude that f ( x ) ≈ − x for | x | sufficiently large. As such, there is no constant L such that lim x →∞ f ( x ) = L or lim x →−∞ f ( x ) = L . Therefore, f ( x ) has no horizontal asymptotes.
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