hw3solutions

# hw3solutions - Math 2250 Written HW#3 Solutions 1 At what...

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Unformatted text preview: Math 2250 Written HW #3 Solutions 1. At what values of x is the function f ( x ) = x tan x x 2 +1 continuous? Answer: It’s most convenient to think of f ( x ) as the product of the functions g ( x ) = x x 2 +1 and h ( x ) = tan x : f ( x ) = x tan x x 2 + 1 = x x 2 + 1 tan x = g ( x ) h ( x ) . Now, the function g ( x ) has continuous numerator and denominator, and the denominator is never zero (since x 2 + 1 ≥ 1 for all x ), so, by the Limit Laws, g ( x ) is continuous everywhere. On the other hand, h ( x ) = tan x = sin x cos x . Both sin x and cos x are continuous everywhere, so h ( x ) will be continuous wherever the denominator is not zero. The function cos x is equal to zero for x = . . . ,- 3 π/ 2 ,- π/ 2 , π/ 2 , 3 π/ 2 , 5 π/ 2 , . . . . In other words, cos x equals zero whenever x = (2 k +1) π 2 for any integer k . Therefore, by the Limit Laws, the function h ( x ) is continuous whenever x is not equal to (2 k +1) π 2 ....
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hw3solutions - Math 2250 Written HW#3 Solutions 1 At what...

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