Explanation Given: The function is f ( x ) = − 4 x 1 + 2 x . Definition used: (1) If the function gets larger and larger in magnitude without bound as x approaches the number k, then the line x = k is the vertical asymptote. (2) A function f ( x ) said to be symmetric about the origin if f ( − x ) = − f ( x ) . Calculation: Domain: The given function f ( x ) = − 4 x 1 + 2 x is defined at all the real number except the point x = − 1 2 . Thus, the domain of the function f ( x ) = − 4 x 1 + 2 x is ( − ∞ , − 1 2 ) ∪ ( − 1 2 , ∞ ) . Symmetry: The given function is f ( x ) = − 4 x 1 + 2 x . Substitute x = − x in f ( x ) , f ( − x ) = − 4 ( − x ) 1 + 2 ( − x ) = 4 x 1 − 2 x Thus, f ( − x ) ≠ − f ( x ) . Therefore, from the definition stated above, the function does not have symmetry about the origin. Intercept: The function f ( x ) intercepts y -axis at x = 0 . Substitute the value x = 0 in f ( x ) , f ( 0 ) = − 4 ( 0 ) 1 + 2 ( 0 ) = 0 That is the function f ( x ) intercepts y -axis at the point ( 0 , 0 ) . The function f ( x ) intercepts x -axis at f ( x ) = 0 . − 4 x 1 + 2 x = 0 − 4 x = 0 x = 0 Thus, x = 0 . That is the function f ( x ) intercepts x -axis at the points ( 0 , 0 ) . Asymptote: Observe that the graph of the function f ( x ) = − 4 x 1 + 2 x tends to the line x = − 1 2 as the value of x tends to − 1 2 . Thus x = − 1 2 is the vertical asymptote for the function f ( x ) = − 4 x 1 + 2 x . Observe the limit, lim x → ∞ f ( x ) = lim x → ∞ − 4 x 1 + 2 x = lim x → ∞ − 4 1 x + 2 = − 4 2 = − 2 . Thus, as the value of x tends to infinity, the graph of the function f ( x ) = − 4 x 1 + 2 x tends to the line y = − 2 . Thus y = − 2 is the horizontal asymptote for the function f ( x ) = − 4 x 1 + 2 x . Critical point: The given function is f ( x ) = − 4 x 1 + 2 x . Differentiate f ( x ) with respect to x , f ′ ( x ) = − 4 ( 1 + 2 x ) − 2 ( − 4 x ) ( 1 + 2 x ) 2 = − 4 − 8 x + 8 x ( 1 + 2 x ) 2 = − 4 ( 1 + 2 x ) 2 Observe that the function f ′ ( x ) = − 4 ( 1 + 2 x ) 2 is never zero

Finite Mathematics and Calculus with Applications (10th Edition)

10th Edition

ISBN: 9780321979407

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 13, Problem 52RE

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Chapter 13, Problem 51RE

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Find all values x = a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a. Be sure to note when the limit doesn't exist. f(x)=4x²+7x+1 Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. (Use a comma to separate answers as needed.) OA. f is discontinuous at the single value x = B. f is discontinuous at the single value x = OC. f is discontinuous at the two values x = OD. fis discontinuous at the two values x = OE. f is discontinuous at the two values x = The limit is The limit does not exist and is not co or - oo. The limit for the smaller value is The limit for the larger value is The limit for both values do not exist and are not co or - co. The limit for the smaller value does not exist and is not oo or - co. The limit for the larger value is

Find all values x = a where the function is discontinuous. For each value of x, give the limit of the function as x approaches a. Be sure to note when the limit doesn't exist. 8+x f(x) = x(x-1) (Use a comma to separate answers as needed.) OA. The function f is discontinuous at the single value x = OB. The function f is discontinuous at the single value x = OC. The function f is discontinuous at the two values x = OD. The function f is discontinuous at the two values x = not oo or -0. OE. The function f is discontinuous at the two values x = The limit is The limit does not exist and is not oo or - co. The limits for both values do not exist and are not co or - co. The limit for the smaller value is The limit for the larger value does not exist and is The limit for the smaller value does not exist and is not co or - co. The limit for the larger

Chapter 13 Solutions

Finite Mathematics and Calculus with Applications (10th Edition)

Ch. 13.1 - Find where the function is increasing and...Ch. 13.1 - Prob. 2YT Ch. 13.1 - Prob. 3YT Ch. 13.1 - Prob. 4YT Ch. 13.1 - Prob. 1WE Ch. 13.1 - Prob. 2WE Ch. 13.1 - Prob. 3WE Ch. 13.1 - Prob. 4WE Ch. 13.1 - Prob. 5WE Ch. 13.1 - Prob. 6WE

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Solution Summary: The author explains that the function f(x) is symmetric about the origin if it gets larger and larger in magnitude as x approaches the number k.

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Chapter 13 Solutions