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10. [ 2 2.7, 6.7] by [ 2 6, 6] A graph of f ( x ) is shown. The range of f ( x ) is ( 2‘ , 1) < [2, ). The values of c for which f ( x ) 5 c has no solution are the values that are excluded from the range. Therefore, c can be any value in [1, 2). Section 2.3 Exercises 1. The function y 5 } ( x 1 1 2) 2 } is continuous because it is a quotient of polynomials, which are continuous. Its only point of discontinuity occurs where it is undefined. There is an infinite discontinuity at x 52 2. 2. The function y 5 } x 2 2 x 1 4 x 1 1 3 } is continuous because it is a quotient of polynomials, which are continuous. Its only points of discontinuity occur where it is undefined, that is, where the denominator x 2 2 4 x 1 3 5 ( x 2 1)( x 2 3) is zero. There are infinite discontinuities at x 5 1 and at x 5 3. 3. The function y 5 } x 2 1 1 1 } is continuous because it is a quotient of polynomials, which are continuous. Furthermore, the domain is all real numbers because the denominator, x 2 1 1, is never zero. Since the function is continuous and has domain ( 2‘ , ), there are no points of discontinuity. 4. The function y 5 ) x 2 1 ) is a composition ( f + g )( x ) of the continuous functions f ( x ) 5 ) x ) and g ( x ) 5 x 2 1, so it is continuous. Since the function is continuous and has domain ( 2‘ , ), there are no points of discontinuity. 5. The function y 5 ˇ 2 w x w 1 w 3 w is a composition ( f + g )( x ) of the continuous functions f ( x ) 5 ˇ x w and g ( x ) 5 2 x 1 3, so it is continuous. Its points of discontinuity are the points not in the domain, i.e., all x ,2} 3 2 } . 6. The function y 5 ˇ 3 2 w x w 2 w 1 w is a composition ( f + g )( x ) of the continuous functions f ( x ) 5 ˇ 3 x w and g ( x ) 5 2 x 2 1, so it is continuous. Since the function is continuous and has domain ( 2‘ , ), there are no points of discontinuity. 7. The function y 5 } ) x x ) } is equivalent to y 5 h It has a jump discontinuity at x 5 0. 8. The function y 5 cot x is equivalent to y 5 } c s o in s x x } , a quotient of continuous functions, so it is continuous. Its only points of discontinuity occur where it is undefined. It has infinite discontinuities at x 5 k p for all integers k . 9. The function y 5 e 1/ x is a composition ( f + g )( x ) of the continuous functions f ( x ) 5 e x and g ( x ) 5 } 1 x } , so it is continuous. Its only point of discontinuity occurs at x 5 0, where it is undefined. Since lim x 0 1 e 1/ x 5‘ , this may be considered an infinite discontinuity. 10. The function y 5 ln ( x 1 1) is a composition ( f + g )( x ) of the continuous functions f ( x ) 5 ln x and g ( x ) 5 x 1 1, so it is continuous. Its points of discontinuity are the points not in the domain, i.e., x ,2 1.

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