Explanation 1) Concept: i) A domain is the set of x values that satisfy the function. ii) To find x-intercept, put y = 0 , and to find y-intercept, put x = 0 in the given function. iii) Symmetry: To find symmetry, replace x by –x and check the behaviour of function. Thus, if f - x = f x , then it is an even function, so it has y -axis symmetry. If f - x = - f x , then it is an odd function, so it has x -axis symmetry. And if f - x ≠ - f x ≠ f x , then it has no symmetry. iv) An asymptote is a tangent at infinity. To find horizontal, vertical, and slant asymptote, follow the rules. v) A function is increasing if f ' x > 0 and decreasing if f ' x < 0 in that particular interval. vi) The number f ( c ) is a local maximum value of f if f c ≥ f ( x ) when x is near c and is a local minimum value of f if f c ≤ f ( x ) when x is near c. vii) If f ' ' x > 0 , the function is concave up and if f ' ' x < 0 , the function is concave down in that particular interval. And if f ' ' x = 0 , give the values of inflection points. 2) Given: y = x x 3 - 1 3) Calculation: Here, first find the domain of the given function and the x & y intercepts. Next, check the symmetry, asymptotes, intervals of increase and decrease, local maximum and minimum values, concavity, and points of inflection. Using these, sketch the curve. A) Domain Since y = x x 3 - 1 is a rational expression, the denominator will be zero at x = 1 . Therefore, the domain is - ∞ , 1 ∪ ( 1 , ∞ ) B) Intercepts For y intercept, plug x = 0 in the given function, and solve it for y . y = 0 0 - 1 y = 0 y intercept is 0 , 0 . For x intercept, plug y = 0 in the original function, and solve it for x . 0 = x x 3 - 1 x = 0 . x Intercept is ( 0 , 0 ) . C) Symmetry For symmetry, replace each x by - x . f - x = - x ( - x ) 3 - 1 = - x - x 3 - 1 = x x 3 + 1 ⇒ f - x ≠ - f ( x ) ≠ f ( x ) Function is neither even nor odd. Therefore, no symmetry. D) Asymptote Horizontal asymptotes lim x → ∞ - ⁡ y = l i m x → ∞ - x x 3 - 1 = 0 , lim x → ∞ + ⁡ y = l i m x → ∞ + x x 3 - 1 = 0 Horizontal asymptote is y = 0 . Vertical asymptotes f x = x x 3 - 1 The denominator become zero when x = 1 , so it has vertical asymptote at x = 1 . Slant asymptotes No slant asymptote. E) Intervals of increase or decrease To find the intervals of increase or decrease, find the derivative of the given function. f ' x = - 2 x 3 - 1 x 3 - 1 2 Equating this derivative with 0 , - 2 x 3 - 1 x 3 - 1 2 = 0 Solving this for x , it becomes x = - 1 2 3 , This is the critical point. By using this critical point and the domain, create intervals. The intervals are - ∞ , - 1 2 3 , - 1 2 3 , 1 , 1 , ∞ . Now, take a test point from each of the above intervals, and check whether the function is increasing or decreasing in that interval. For - ∞ , - 1 2 3 , consider x = - 1 . f ' - 1 = - 2 ( - 1 ) 3 - 1 ( - 1 ) 3 - 1 2 f ' - 1 = 1 4 - ∞ < x < - 1 2 3 ; f ' x > 0 The function is increasing in the interval ( - ∞ , - 1 2 3 ) . For - 1 2 3 , 1 , consider x = 0 . f ' 0 = - 2 ( 0 ) 3 - 1 ( 0 ) 3 - 1 2 f ' 0 = - 1 - 1 2 3 < x < 1 ; f ' x > 0 The function is increasing in the interval ( - 1 2 3 , 1 )

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ISBN: 9781337051545

Author: Stewart

Publisher: CENGAGE C

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Chapter 3.5, Problem 18E

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The curve of the given function

Solution Summary: The author explains that a domain is the set of x values that satisfy the function.

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To sketch:

The curve of the given function