Explanation Domain: The given polynomial is f ( x ) = 27 ( x − 2 ) 2 ( x + 2 ) . Notice that the given polynomial is defined on all real numbers. Hence, the domain of f is ( − ∞ , ∞ ) . Symmetry: The function is even if f ( − x ) = f ( x ) and odd if f ( − x ) = − f ( x ) . Thus, find the value of f ( − x ) . f ( − x ) = 27 ( − x − 2 ) 2 ( − x + 2 ) = − 27 ( x + 2 ) 2 ( x − 2 ) Notice that, f ( − x ) ≠ f ( x ) and f ( − x ) ≠ − f ( x ) . This is neither odd symmetry nor even symmetry. Therefore, the polynomial has no symmetry. Derivatives: Rewrite the given function as, f ( x ) = 27 ( x − 2 ) 2 ( x + 2 ) = 27 ( x 2 + 4 − 4 x ) ( x + 2 ) = 27 ( x 3 + 2 x 2 + 4 x + 8 − 4 x 2 − 8 x ) = 27 x 3 − 54 x 2 − 108 x + 216 . Differentiate f ( x ) = 27 x 3 − 54 x 2 − 108 x + 216 with respect to x on both sides, d d x ( f ( x ) ) = d d x ( 27 x 3 − 54 x 2 − 108 x + 216 ) f ′ ( x ) = 81 x 2 − 108 x − 108 Differentiate f ′ ( x ) = 81 x 2 − 108 x − 108 with respect to x on both sides, d d x ( f ′ ( x ) ) = d d x ( 81 x 2 − 108 x − 108 ) = 162 x − 108 Critical points and possible inflections points: Set first derivative to 0 and find the critical points. f ′ ( x ) = 0 81 x 2 − 108 x − 108 = 0 User the quadratic Formula − b ± b 2 − 2 a c 2 a to calculate the two roots of the equation 81 x 2 − 108 x − 108 = 0 as follows. x = − ( − 108 ) ± ( − 108 ) 2 − 4 × 81 × ( − 108 ) 2 × 81 = 108 ± 46656 162 = 108 ± 216 162 = 2 , − 2 3 Thus, the critical points are at x = 2 and x = − 2 3 . The corresponding interval to the critical points is ( − ∞ , − 2 3 ) ∪ ( − 2 3 , 2 ) ∪ ( 2 , ∞ ) . Substitute x = 2 in f ( x ) = 27 x 3 − 54 x 2 − 108 x + 216 , f ( 2 ) = 27 × 2 3 − 54 × 2 2 − 108 × 2 + 216 = 216 − 216 − 216 + 216 = 0 Substitute x = − 2 3 in f ( x ) = 27 x 3 − 54 x 2 − 108 x + 216 , f ( − 2 3 ) = 27 × ( − 2 3 ) 3 − 54 × ( − 2 3 ) 2 − 108 × ( − 2 3 ) + 216 = − 8 − 36 + 72 + 216 = 244 Thus, the critical points are ( 2 , 0 ) and ( − 2 3 , 244 ) . Set second derivative to 0 and find the inflection points. f ′ ′ ( x ) = 0 162 x − 108 = 0 x = 108 162 = 2 3 Substitute x = 2 3 in f ( x ) = 27 x 3 − 54 x 2 − 108 x + 216 as, f ( 2 3 ) = 27 × ( 2 3 ) 3 − 54 × ( 2 3 ) 2 − 108 × ( 2 3 ) + 216 = 8 − 36 + 72 + 216 = 260 Thus, the possible inflection point is ( 2 3 , 260 )

Calculus, Single Variable: Early Transcendentals (3rd Edition)

3rd Edition

ISBN: 9780134766850

Author: William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz

Publisher: PEARSON

See similar textbooks

Concept explainers

Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.

Videos

Question

Chapter 4.4, Problem 22E

To determine

To sketch: The graph of the polynomial f(x)=27(x−2)2(x+2) and make a complete graph of f.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 4.4, Problem 21E

Chapter 4.4, Problem 23E

Chapter 4 Solutions

Calculus, Single Variable: Early Transcendentals (3rd Edition)

Ch. 4.1 - Sketch the graph of a function that is continuous...Ch. 4.1 - Consider the function f(x) = x3. Where is the...Ch. 4.1 - Sketch the graph of a function that is continuous...Ch. 4.1 - Sketch the graph of a function that has an...Ch. 4.1 - What is a critical point of a function?Ch. 4.1 - Sketch the graph of a function f that has a local...Ch. 4.1 - Sketch the graph of a function f that has a local...Ch. 4.1 - Absolute maximum/minimum values Use the following...Ch. 4.1 - Prob. 12E Ch. 4.1 - Absolute maximum/minimum values Use the following...

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.

The graph of the polynomial f ( x ) = 27 ( x − 2 ) 2 ( x + 2 ) and make a complete graph of f .

Solution Summary: The author sketched the graph of the polynomial f(x)=27 x 22 left (x+2).

Concept explainers

Videos

To sketch: The graph of the polynomial f(x)=27(x−2)2(x+2) and make a complete graph of f.

Want to see the full answer?

Chapter 4 Solutions