Explanation Given: The equation of the curve is as below. f ( x ) = ( x 4 + 1 ) x (1) Calculation: Apply the condition lim x → ∞ [ f ( x ) − x 3 ] = 0 . lim x → ∞ [ f ( x ) − x 3 ] = lim x → ∞ ( x 4 + 1 x − x 3 ) = lim x → ∞ ( x 4 + 1 − x 4 x ) = lim x → ∞ 1 x = 0 Therefore, the condition lim x → ∞ [ f ( x ) − x 3 ] = 0 is true. Clearly the curve y = f ( x ) is asymptotic to y = x 3 . Compute the domain. Consider the denominator in equation (1) x = 0 Therefore, the domain is ( − ∞ , 0 ) ∪ ( 0 , ∞ ) . Compute the y -intercepts. Substitute the value 0 for x in equation (1). f ( 0 ) = ( 0 4 + 1 ) 0 f ( 0 ) = ∞ Compute x intercepts. Substitute the value 0 for f ( x ) in the numerator of equation (1). f ( x ) = 0 x 4 + 1 = 0 x 4 = 1 x = 1 4 So the equation has no real solution. Therefore, there are no x and y intercepts. Compute the symmetric. Apply the negative and positive values for x in equation (1). Substitute the value of − 1 for x . f ( − 1 ) = ( − 1 4 + 1 ) − 1 f ( − 1 ) = 2 − 1 f ( − 1 ) = − 2 Substitute the value of 1 for x . f ( 1 ) = ( 1 4 + 1 ) 1 f ( 1 ) = 2 1 f ( 1 ) = 2 f ( − x ) = − f ( x ) ≠ f ( x ) So the function is odd. Therefore, the curve is symmetric about the origin. Compute the vertical asymptote. lim x → 0 + ( x 4 + 1 ) x = ∞ lim x → 0 − ( x 4 + 1 ) x = − ∞ Therefore, the vertical asymptote is x = 0 _ . Compute the horizontal asymptote. In equation (1), the value of numerator is greater than the denominator, and so the curve will not have horizontal asymptote. Then, the curve y = f ( x ) is asymptotic to the parabola y = x 3 . Compute the interval of increase or decrease. Differentiate equation (1) as follows. f ' ( x ) = d d x ( ( x 4 + 1 ) x ) f ' ( x ) = x ( 4 x 3 + 0 ) − ( x 4 + 1 ) x 2 f ' ( x ) = 3 x 4 − 1 x 2 Check whether the condition f ' ( x ) > 0 is true inside the limit − ∞ < x < 1 3 4 . Substitute the value − 2 for x in the above equation. f ' ( − 2 ) = 3 ( − 2 ) 4 − 1 ( − 2 ) 2 f ' ( − 2 ) = 11.75 Hence, the condition f ' ( x ) > 0 is true. Therefore, the function f ( x ) is increasing on ( − ∞ , 1 3 4 ) . Check whether the condition f ' ( x ) > 0 is true in the limit 1 3 4 < x < ∞ . Substitute the value 2 for x in the above equation. f ' ( 2 ) = 3 ( 2 ) 4 − 1 ( 2 ) 2 f ' ( 2 ) = 11.75 Hence, the condition f ' ( x ) > 0 is true. Therefore, the function f ( x ) is increasing on ( 1 3 4 , ∞ ) . Check whether the condition f ' ( x ) < 0 is true in the limit − 1 3 4 < x < 0 . Substitute the value − 0.5 for in the above equation. f ' ( − 0.5 ) = 3 ( − 0.5 ) 4 − 1 ( − 0.5 ) 2 f ' ( − 0.5 ) = − 3.25 < 0 Hence, the condition f ' ( x ) < 0 is true. Therefore, the function f ( x ) is decreasing inside the domain ( − 1 3 4 , 0 ) . Check whether the condition f ' ( x ) < 0 is true in the limit 0 < x < 1 3 4

Single Variable Calculus: Early Transcendentals

8th Edition

ISBN: 9781305748217

Author: James Stewart

Publisher: CENGAGE C

See similar textbooks

Concept explainers

Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

Slope

The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

Videos

Question

Chapter 4.5, Problem 75E

To determine

To show: Applying the expression limx→∞[f(x)−x3]=0 shows that the graph of f approaches the graph of y=x3 and y=f(x) is asymptotic to the parabola y=x3 and to sketch the curve.

Expert Solution & Answer

Chapter 4 Solutions

Single Variable Calculus: Early Transcendentals

Ch. 4.1 - Explain the difference between an absolute minimum...Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - For each of the numbers a, b, c, d, r, and s,...Ch. 4.1 - Prob. 5E Ch. 4.1 - Use the graph to state the absolute and local...Ch. 4.1 - Prob. 7E Ch. 4.1 - Sketch the graph of a function f that is...Ch. 4.1 - Sketch the graph of a function f that is...Ch. 4.1 - Prob. 10E

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

Applying the expression lim x → ∞ [ f ( x ) − x 3 ] = 0 shows that the graph of f approaches the graph of y = x 3 and y = f ( x ) is asymptotic to the parabola y = x 3 and to sketch the curve.

Solution Summary: The author illustrates how the expression undersetxtoinftymathrmlimleft is asymptotic to the parabola and to sketch the curve.

Concept explainers

Videos

To show: Applying the expression limx→∞[f(x)−x3]=0 shows that the graph of f approaches the graph of y=x3 and y=f(x) is asymptotic to the parabola y=x3 and to sketch the curve.

Chapter 4 Solutions