This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 nonnegative 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 ≥ . 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 − = f ( x ) x = f ′ ( c ) for some c ∈ (0 , x ) . But f ′ ( c ) = 0 which implies that f ( x ) = 0. 1 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 xaxis 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 1 2 3 4 5 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)....
View
Full
Document
This note was uploaded on 12/31/2011 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.
 Fall '08
 ANDREWCHILDS
 Math, Algebra

Click to edit the document details