Explanation Given data: y The given functions are: (i) g ( x ) = 1 2 x − 1 / 2 ( 1 + 5 x ) − x ( 5 ) ( 1 + 5 x ) 2 , (ii) g ( x ) = 2 x x − 1 − x 2 ( 1 2 ( x − 1 ) − 1 / 2 ) (iii) g ( x ) = e x ( 1 − e x ) − e x ( − e x ) ( 1 − e x ) 2 Method used: The solution/s of a function are the values of “ x ” for which g ( x ) = 0 , and function g ( x ) does not exist at the point that makes it indefinite. Calculation: (i).The solutions of function g ( x ) = 1 2 x − 1 / 2 ( 1 + 5 x ) − x ( 5 ) ( 1 + 5 x ) 2 , and the point where it doesnot exist. First, we simplify the function as: g ( x ) = ( 1 + 5 x ) 2 x − x ( 5 ) ( 1 + 5 x ) 2 = ( 1 + 5 x ) − 2 × 5 x x 2 x ( 1 + 5 x ) 2 = 1 + 5 x − 10 x 2 x ( 1 + 5 x ) 2 = 1 − 5 x 2 x ( 1 + 5 x ) 2 The solution of function g ( x ) are the values of x for which g ( x ) = 0 1 − 5 x 2 x ( 1 + 5 x ) 2 = 0 1 − 5 x = 0 x = 1 5 Now, the function g ( x ) = 1 − 5 x 2 x ( 1 + 5 x ) 2 . Clearly, the function is undefined when 2 x ( 1 + 5 x ) 2 = 0 x = 0 , x = − 1 5 Thus, solution of the function g ( x ) = 1 2 x − 1 / 2 ( 1 + 5 x ) − x ( 5 ) ( 1 + 5 x ) 2 is x = 1 5 , and function does not exist at points x = 0 and x = − 1 5 . (ii) simply function g ( x ) = 2 x x − 1 − x 2 ( 1 2 ( x − 1 ) − 1 / 2 )

14th Edition

ISBN: 9781464151088

Author: Associate Professor Laura Taalman, Peter Kohn

Publisher: W.H. Freeman & Company

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8:38 *** TEMU TEMU -3 -2 7 B 2 1 & 5G. 61% 1 2 -1 Based on the graph above, determine the amplitude, period, midline, and equation of the function. Use f(x) as the output. Amplitude: 2 Period: 2 Midline: 2 ☑ syntax error: this is not an equation. Function: f(x) = −2 cos(πx + 2.5π) +2× Question Help: Worked Example 1 ☑ Message instructor Submit Question ||| <

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The solutions of given functions and the points where the functions do not exist.

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Chapter 3 Solutions