CK-12 Calculus Study Guide - Even and Odd Functions

Sanjeev Narayanaswamy

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Printed: April 6, 2015 www.ck12.org Chapter 1. CK-12 Calculus Study Guide - Even and Odd Functions

CHAPTER 1 CK-12 Calculus Study Guide - Even and Odd Functions

Important Points - Even and Odd Functions

Let f (x) represent any function. f (x) can be classiﬁed as either even, odd, or neither. Even Function

• f (x) is deﬁned to be an even function if and only if it satisﬁes f (−x) = f (x) • f (x) is symmetric about the y axis. i.e. f (x) = x2

Odd Function

• f (x) is deﬁned to be an odd function if it satisﬁes f (−x) = − f (x) • f (x) is rotationally symmetric about the origin. i.e. f (x) = x3

Neither Odd nor Even

• A function is neither odd nor even if it doesn’t satisfy the conditions for odd or even functions. • A quick test for this is if f (−x) 6= f (x) and f (−x) 6= − f (x).

Useful Properties

• sin(−x) = −sin(x) • cos(−x) = cos(x) • |−x|= |x| • If f (x) = xn, where n is a natural number (positive, non-zero integer), then f (x) is even if n is even and odd if n is odd.

Example 1

Question: Which of the following is NOT a property of an even function? Pick the best answer. a) symmetric about y axis b) f (x) = − f (x) c) symmetric about x axis d) a and b e) b and c f) none of the above

Analysis:

An even function must satisfy f (−x) = f (x).This also implies symmetry about the y-axis. From the options given, the properties listed in b) and c) are both not properties of even functions. Answer: e)

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Example 2

Question: Classify the following function as even, odd, or neither:

f (x) = (2x2 + 5x)2

Analysis:

The ﬁrst step in the algebraic approach to classifying functions as even, odd, or neither is to replace x with −x. Then we need to see if f (−x) = f (x) or f (−x) = − f (x) is satisﬁed.

f (x) = (2x2 + 5x)2 f (−x) = (2(−x)2 + 5(−x))2 = (2(−x)2 + 5(−x))2 = (2(x)2 − 5(x))2 = (2x2 − 5x)2 − f (x) = −(2x2 + 5x)2

In this analysis, we see that f (−x) 6= − f (x) and f (−x) 6= f (x). Answer: Neither Below is the graph of the function for reference:

Example 3

Question: Classify the following function as even, odd, or neither:

|x| f (x) = (|2x|+1)

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Analysis: The ﬁrst step in the algebraic approach to classifying functions as even, odd, or neither is to replace x with −x. Then we need to see if f (−x) = f (x) or f (−x) = − f (x) is satisﬁed.

|x| f (x) = (|2x|+1) |−x| f (−x) = (|2(−x)|+1) |−x| = (|−2x|+1) |x| = (|2x|+1) In this analysis, we have used the property that |x|= |−x|. We see that f (x) = f (−x). Answer: Even Below is the graph of the function for reference:

Example 4

Question: Classify the following function as even, odd, or neither:

f (x) = |sin(x)|

Analysis: Check if f (−x) = f (x) or f (−x) = − f (x) is satisﬁed.

f (x) = |sin(x)| f (−x) = |sin(−x)| = |−sin(x)| = |sin(x)|

In this analysis, we have used the property that sin(−x) = −sin(x) and |x|= |−x|. We see that f (x) = f (−x). Answer: Even Below is the graph of the function for reference:

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Example 5

Question: Classify the following function as even, odd, or neither:

|sin(x)| f (x) = sin(x)

Analysis: Check if f (−x) = f (x) or f (−x) = − f (x) is satisﬁed.

In this analysis, we have used the property that sin(−x) = −sin(x) and |x|= |−x|. We see that f (−x) = − f (x). Answer: Odd Below is the graph of the function for reference:

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Note that the given function is not deﬁned for x = nπ, where n is any integer.

Example 6

Question: Classify the following function as even, odd, or neither:

f (x) = ex

Analysis: Check if f (−x) = f (x) or f (−x) = − f (x) is satisﬁed.

f (x) = ex f (−x) = e−x

In this analysis, we see that f (−x) 6= − f (x) and f (−x) 6= f (x). Answer: Neither Below is the graph of the function for reference:

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Example 7

Question: Classify the following function as even, odd, or neither:

q f (x) = |x3|

Analysis: Check if f (−x) = f (x) or f (−x) = − f (x) is satisﬁed.

q f (x) = |x3| q f (−x) = |(−x)3| q = |(−1)3x3| q = |−x3| q = |x3|

In this analysis, we have used the property that |x|= |−x|. We see that f (x) = f (−x). Answer: Even Below is the graph of the function for reference:

Example 8

Question: Classify the following function as even, odd, or neither:

f (x) = cos(x) + sin(x)

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Analysis: Check if f (−x) = f (x) or f (−x) = − f (x) is satisﬁed.

f (x) = cos(x) + sin(x) f (−x) = cos(−x) + sin(−x) = cos(−x) + sin(−x) = cos(x) + (−sin(x))

In this analysis, we have used the property that cos(−x) = cos(x) and sin(−x) = −sin(x). We see that f (−x) 6= − f (x) and f (−x) 6= f (x). Answer: Neither Below is the graph of the function for reference:

Example 9

Question: Classify the following function as even, odd, or neither:

f (x) = sin[sin(x)]

Analysis: Check if f (−x) = f (x) or f (−x) = − f (x) is satisﬁed.

f (x) = sin[sin(x)] f (−x) = sin[sin(−x)] = sin[−sin(x)] = −sin[sin(x)]

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In this analysis, we have used the property that sin(−x) = −sin(x) twice to bring out the negative sign. We see that f (−x) = − f (x). Answer: Odd Below is the graph of the function for reference:

Example 10

Question: Classify the following function as even, odd, or neither:

|x| sin(ee ) f (x) = pcos(x)

Analysis: Check if f (−x) = f (x) or f (−x) = − f (x) is satisﬁed.

|x| sin(ee ) f (x) = pcos(x) |−x| sin(ee ) f (−x) = pcos(−x) |x| sin(ee ) = pcos(x)

To analyze this very complicated function, we have simply used the property that |x|= |−x| and cos(−x) = cos(x). We see that f (−x) = − f (x). Answer: Even Below is the graph of the function for reference:

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