Explanation The given function is C = 0.15 t t 2 + 1 . It is known that the points for which the graph crosses each axis is called the intercepts of the graph. Set c = 0 and obtain the t- intercept of the graph as follows: 0.15 t t 2 + 1 = 0 Since the above equation is satisfied for the value of t , the curve C = 0.15 t t 2 + 1 have x- intercept at origin. Set t = 0 in the curve C = 0.15 t t 2 + 1 and obtain the y- intercept of the graph as follows: Thus, the curve C = 0.15 t t 2 + 1 have y- intercept at origin. Obtain the symmetry of the given function as follows. Substitute t = − t in the given equation C = − ( 0.15 t t 2 + 1 ) and check the symmetry of the given function as follows: Here, the curve C = 0.15 t t 2 + 1 is not same as = − 0.15 t t 2 + 1 . Hence, the curve C = 0.15 t t 2 + 1 is not symmetric about y -axis. It is to be noted that the curve C = 0.15 t t 2 + 1 is not same as − C = 0.15 t t 2 + 1 . Thus, the curve C = 0.15 t t 2 + 1 is not symmetric about x -axis. Obtain the asymptotes of the given function as follows. As t → ∞ or t → − ∞ , the curve C = 0.15 t t 2 + 1 tends to the graph of c → ∞ . Thus, there are no horizontal asymptote of the graph because at c = 0 then t = 0 . Since the function C = 0.15 t t 2 + 1 is defined for t = 0 , but t → 0 then c → 0 so there are no vertical asymptote of the graph. It is to be noted that, the function C = 0.15 t t 2 + 1 is defined for all real numbers. Thus, the domain of the function is ( − ∞ , ∞ ) . The function C = 0.15 t t 2 + 1 attains positive and negative values. Thus, the range of the function C = 0.15 t t 2 + 1 is [ − 0.075 , 0.075 ] . Differentiate the equation C = 0.15 t t 2 + 1 with respect to t and obtain the first derivative as follows: d c d x = d d x ( 0.15 t t 2 + 1 ) = 0

Student Solutions Manual For Basic Technical Mathematics And Basic Technical Mathematics With Calculus

11th Edition

ISBN: 9780134434636

Author: Allyn J. Washington, Richard Evans

Publisher: PEARSON

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Question

Chapter 24.6, Problem 24E

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To sketch: The graph of the function C=0.15tt2+1.

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Chapter 24.6, Problem 23E

Chapter 24.6, Problem 25E

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