Explanation Given Information: The provided function is, y = x 2 − 3 x + 6 x − 2 Formula used: A vertical asymptote of a function f ( x ) is a line x = a such that f ( a ) = ∞ . A vertical asymptote of a rational function h ( x ) = f ( x ) g ( x ) is x = a where g ( a ) = 0 and f ( a ) ≠ 0 . A horizontal asymptote of a function f ( x ) is a line y = b such that lim x → ∞ f ( x ) = b or lim x → − ∞ f ( x ) = b . A horizontal asymptote of a rational function h ( x ) = f ( x ) g ( x ) is 1. A line y = 0 if the degree of the numerator is less than the degree of the denominator. 2. The line y is ratio of the leading coefficients if the degree of the numerator is equal to the degree of the denominator. 3. Does not exist if the degree of the numerator is greater than the degree of the denominator. To find relative maxima and minima of a function, 1. Set the first derivative of the function to zero, f ′ ( x ) = 0 , to find the critical values of the function. 2. Substitute the critical values into f ( x ) and calculate the critical points. 3. Calculate the sign of the function to the left and right of the critical values. (a) If f ′ ( x ) < 0 and f ′ ( x ) > 0 on the left and right side respectively of a critical value, then the point is relative minimum. (b) If f ′ ( x ) > 0 and f ′ ( x ) < 0 on the left and right side respectively of a critical value, then the point is relative maximum. Calculation: Consider the provided function, y = x 2 − 3 x + 6 x − 2 Now, calculate the first derivative. y ′ = ( x − 2 ) ( 2 x − 3 ) − ( x 2 − 3 x + 6 ) 1 ( x − 2 ) 2 = 2 x 2 − 3 x − 4 x + 6 − x 2 + 3 x − 6 ( x − 2 ) 2 = x ( x − 4 ) ( x − 2 ) 2 Now, to obtain the critical values, set y ′ = 0 as, x ( x − 4 ) ( x − 2 ) 2 = 0 x ( x − 4 ) = 0 Thus, either ( x − 4 ) = 0 or x = 0 . First consider ( x − 4 ) = 0 . Add 4 to both sides and simplify. x − 4 + 4 = 0 + 4 x = 4 Thus, the critical values of the function are at x = 0 and x = 4 . Now, substitute 0 for x in the equation y = x 2 − 3 x + 6 x − 2 , y = ( 0 ) 2 − 3 ( 0 ) + 6 0 − 2 = 6 − 2 = − 3 Substitute 4 for x in the equation y = x 2 − 3 x + 6 x − 2 , y = ( 4 ) 2 − 3 ( 4 ) + 6 4 − 2 = 16 − 12 + 6 2 = 10 2 = 5 Thus, the critical points are ( 0 , − 3 ) and ( 4 , 5 ) . Now, calculate the sign of the y ′ on either side of the critical point. First consider x = 0 . Calculate y ′ at x = − 1 . y ′ = x ( x − 4 ) ( x − 2 ) 2 = − 1 ( − 1 − 4 ) ( − 1 − 2 ) 2 = − 1 ( − 5 ) ( − 3 ) 2 = 5 9 Thus, y ′ > 0 on left of x = 0

Mathematical Applications for the Management, Life, and Social Sciences

11th Edition

ISBN: 9781305108042

Author: Ronald J. Harshbarger, James J. Reynolds

Publisher: Cengage Learning

See similar textbooks

Concept explainers

Minimization

In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in ter…

Maxima and Minima

The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.

Derivatives

A derivative means a change. Geometrically it can be represented as a line with some steepness. Imagine climbing a mountain which is very steep and 500 meters high. Is it easier to climb? Definitely not! Suppose walking on the road for 500 meters. Which …

Concavity

In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the …

Videos

Question

Chapter 10, Problem 3T

To determine

To calculate: The relative minimum, relative maximum, points of inflection and asymptotes for the provided function y=x2−3x+6x−2 and sketch its graph.

Expert Solution & Answer

Chapter 10 Solutions

Mathematical Applications for the Management, Life, and Social Sciences

Ch. 10.1 - 1. The x- values of critical points are found...Ch. 10.1 - Prob. 2CP Ch. 10.1 - Prob. 3CP Ch. 10.1 - Prob. 4CP Ch. 10.1 - In Problems 1 and 2, use the indicated points on...Ch. 10.1 - In Problems 1 and 2, use the indicated points on...Ch. 10.1 - 3. Use the graph of in Problem 1 to identify at...Ch. 10.1 - 4. Use the graph of in Problem 2 to identify at...Ch. 10.1 - In Problems 5 and 6, use the sign diagram for to...Ch. 10.1 - In Problems 5 and 6, use the sign diagram for to...

Knowledge Booster

Learn more about

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

The relative minimum, relative maximum , points of inflection and asymptotes for the provided function y = x 2 − 3 x + 6 x − 2 and sketch its graph.

Solution Summary: The author calculates the relative minimum, relative maximum, points of inflection and asymptotes for the provided function y=x2-3x+6

Concept explainers

Videos

To calculate: The relative minimum, relative maximum, points of inflection and asymptotes for the provided function y=x2−3x+6x−2 and sketch its graph.

Chapter 10 Solutions