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MATH160Lecture 2.1:
Calculus: Symmetry of Functions
Sarah Wakes
University of Otago
25 August 2021
Symmetries
f is even if f (−x) = f (x) for all x =⇒ graph is symmetric aboutthe y -axis.
f is odd if f (−x) = −f (x) for all x =⇒ graph is symmetric aboutthe origin.
MATH160 - Lecture 2.1: Symmetry 24
Symmetries
f is even if f (−x) = f (x) for all x =⇒ graph is symmetric aboutthe y -axis.
f is odd if f (−x) = −f (x) for all x =⇒ graph is symmetric aboutthe origin.
MATH160 - Lecture 2.1: Symmetry 24
Examples
Discuss the symmetries of the following functions:
f (x) = x2
g(x) =√x2 − x
p(x) =√2− x2
q(x) =1
2x − x3
MATH160 - Lecture 2.1: Symmetry 25
Examples
For f (x) = x2
Substituting −x into the function gives:
f (−x) = (−x)2 = x2 = f (x)
Therefore as f (−x) = f (x) the function is even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For f (x) = x2
Substituting −x into the function gives:
f (−x) = (−x)2 = x2 = f (x)
Therefore as f (−x) = f (x) the function is even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For f (x) = x2
Substituting −x into the function gives:
f (−x) = (−x)2 = x2 = f (x)
Therefore as f (−x) = f (x) the function is even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For f (x) = x2
Substituting −x into the function gives:
f (−x) = (−x)2 = x2 = f (x)
Therefore as f (−x) = f (x) the function is even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For g(x) =√x2 − x
Substituting −x into the function gives:
g(−x) =√
(−x)2 − (−x) =√x2 + x
Therefore as g(x) is neither odd or even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For g(x) =√x2 − x
Substituting −x into the function gives:
g(−x) =√
(−x)2 − (−x) =√x2 + x
Therefore as g(x) is neither odd or even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For g(x) =√x2 − x
Substituting −x into the function gives:
g(−x) =√(−x)2 − (−x) =
√x2 + x
Therefore as g(x) is neither odd or even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For g(x) =√x2 − x
Substituting −x into the function gives:
g(−x) =√(−x)2 − (−x) =
√x2 + x
Therefore as g(x) is neither odd or even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For p(x) =√2− x2
Substituting −x into the function gives:
p(−x) =√2− (−x)2 =
√2− x2 = p(x)
Therefore p(x) is even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For p(x) =√2− x2
Substituting −x into the function gives:
p(−x) =√2− (−x)2 =
√2− x2 = p(x)
Therefore p(x) is even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For p(x) =√2− x2
Substituting −x into the function gives:
p(−x) =√2− (−x)2 =
√2− x2 = p(x)
Therefore p(x) is even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For p(x) =√2− x2
Substituting −x into the function gives:
p(−x) =√2− (−x)2 =
√2− x2 = p(x)
Therefore p(x) is even.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For q(x) = 12x−x3
Substituting −x into the function gives:
q(−x) = 12(−x)−(−x)3
= 1−2x+x3
= 1−(2x−x3)
= −12x−x3
= −q(x)
Therefore q(x) is odd.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For q(x) = 12x−x3
Substituting −x into the function gives:
q(−x) = 12(−x)−(−x)3
= 1−2x+x3
= 1−(2x−x3)
= −12x−x3
= −q(x)
Therefore q(x) is odd.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For q(x) = 12x−x3
Substituting −x into the function gives:
q(−x) = 12(−x)−(−x)3
= 1−2x+x3
= 1−(2x−x3)
= −12x−x3
= −q(x)
Therefore q(x) is odd.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For q(x) = 12x−x3
Substituting −x into the function gives:
q(−x) = 12(−x)−(−x)3
= 1−2x+x3
= 1−(2x−x3)
= −12x−x3
= −q(x)
Therefore q(x) is odd.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For q(x) = 12x−x3
Substituting −x into the function gives:
q(−x) = 12(−x)−(−x)3
= 1−2x+x3
= 1−(2x−x3)
= −12x−x3
= −q(x)
Therefore q(x) is odd.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For q(x) = 12x−x3
Substituting −x into the function gives:
q(−x) = 12(−x)−(−x)3
= 1−2x+x3
= 1−(2x−x3)
= −12x−x3
= −q(x)
Therefore q(x) is odd.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For q(x) = 12x−x3
Substituting −x into the function gives:
q(−x) = 12(−x)−(−x)3
= 1−2x+x3
= 1−(2x−x3)
= −12x−x3
= −q(x)
Therefore q(x) is odd.
MATH160 - Lecture 2.1: Symmetry 25
Examples
For q(x) = 12x−x3
Substituting −x into the function gives:
q(−x) = 12(−x)−(−x)3
= 1−2x+x3
= 1−(2x−x3)
= −12x−x3
= −q(x)
Therefore q(x) is odd.
MATH160 - Lecture 2.1: Symmetry 25
