Explanation Domain: Notice that, the given polynomial f ( x ) = 2 − 2 x 2 3 + x 4 3 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, f ( − x ) = 2 − 2 ( − x ) 2 3 + ( − x ) 4 3 = 2 − 2 x 2 3 + x 4 3 = f ( x ) Notice that, the given function is f ( − x ) = f ( x ) . That is, the function is neither even nor odd function. Hence, the given function is even symmetry. Derivatives: Differentiate f ( x ) = 2 − 2 x 2 3 + x 4 3 with respect to x on both sides as, d d x ( f ( x ) ) = d ( 2 − 2 x 2 3 + x 4 3 ) d x f ′ ( x ) = − 4 3 x − 1 3 + 4 3 x 1 3 = 4 x 2 3 − 4 3 x 1 3 Differentiate f ′ ( x ) = − 4 3 x − 1 3 + 4 3 x 1 3 with respect to x on both sides as, d d x ( f ′ ( x ) ) = d d x ( − 4 3 x − 1 3 + 4 3 x 1 3 ) f ′ ′ ( x ) = ( 4 3 ) × ( 1 3 ) x − 1 3 − 1 + 4 3 × 1 3 x 1 3 − 1 = 4 9 x − 4 3 + 4 9 x − 2 3 Critical points and possible inflection points: Set the first derivative to 0 and find the critical points as follows. f ′ ( x ) = 0 − 4 3 x − 1 3 + 4 3 x 1 3 = 0 4 x 2 3 − 4 3 x 1 3 = 0 This undefined when at x = 0 and is equal to zero when x 2 3 = 1 . More simplification becomes x = ± 1 Thus, the critical points occurs at x = ± 1 . Substitute x = 1 in f ( x ) = 2 − 2 x 2 3 + x 4 3 as, f ( 1 ) = 2 − 2 ( 1 ) 2 3 + ( 1 ) 4 3 = 1 Substitute x = − 1 in f ( x ) = 2 − 2 x 2 3 + x 4 3 as, f ( − 1 ) = 2 − 2 ( − 1 ) 2 3 + ( − 1 ) 4 3 = 1 Thus, the critical points are ( 1 , 1 ) and ( − 1 , 1 ) . The corresponding interval to the critical points is ( − ∞ , − 1 ) ∪ ( − 1 , 1 ) ∪ ( 1 , ∞ ) . Set the second derivative to 0 and find the inflection points as follows. f ′ ′ ( x ) = 0 4 9 x − 4 3 + 4 9 x − 2 3 = 0 4 x 4 3 = − 4 x 2 3 ( 1 x 4 3 ) 3 = ( − 1 x 2 3 ) 3 1 x 4 = − 1 x 2 The above equation has no solution for x ∈ R

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

3rd Edition

ISBN: 9780134766850

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

Publisher: PEARSON

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Chapter 4.4, Problem 40E

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To sketch: The complete graph of the polynomial f(x)=2−2x23+x43.

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Chapter 4.4, Problem 39E

Chapter 4.4, Problem 41E

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...

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The complete graph of the polynomial f ( x ) = 2 − 2 x 2 3 + x 4 3 .

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To sketch: The complete graph of the polynomial f(x)=2−2x23+x43.

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