Lesson 1.4 · Unit 1 · F-BF.3
Testing a function for a mirror line or a half-turn
Some graphs look the same in a mirror, and some look the same after being turned upside down about the origin. Both properties can be tested with algebra alone, and they halve the work in graphing, in integrating later, and in trigonometry, where sine and cosine are opposites in exactly this sense.
- A function is even if \( f(-x) = f(x) \) for every \( x \) in its domain. The graph is symmetric about the \( y \)-axis.
- A function is odd if \( f(-x) = -f(x) \) for every \( x \). The graph is symmetric about the origin: a half-turn maps it onto itself.
- To test, compute \( f(-x) \) by replacing every \( x \) with \( (-x) \), and simplify fully.
- Compare the result with \( f(x) \) and with \( -f(x) \). Equal to the first: even. Equal to the second: odd. Neither: neither.
- Even powers of \( x \) give even terms; odd powers give odd terms. A polynomial with both kinds is usually neither.
- A single numerical example can disprove a symmetry but never prove one. Proving requires the algebra for all \( x \).
- The domain must itself be symmetric. If \( x \) is allowed then \( -x \) must be too.
- An odd function defined at 0 must have \( f(0) = 0 \), because \( f(0) = -f(0) \).
Where students lose marks: concluding a function is neither because \( f(-x) \ne f(x) \), without checking \( f(-x) = -f(x) \). Test both before giving up. For \( f(x) = x^3 \), the first comparison fails and the second succeeds.
The problem. Classify each function as even, odd or neither. (a) \( f(x) = x^4 - 3x^2 \). (b) \( g(x) = x^3 + x \). (c) \( h(x) = x^2 + x \). (d) \( k(x) = \dfrac{x}{x^2 + 1} \). Then state what happens when such functions are multiplied.
Step one: test (a). \( f(-x) = (-x)^4 - 3(-x)^2 = x^4 - 3x^2 \). That equals \( f(x) \) exactly, so \( f \) is even.
Step two: test (b). \( g(-x) = (-x)^3 + (-x) = -x^3 - x \). Compare: \( -g(x) = -(x^3 + x) = -x^3 - x \). They match, so \( g \) is odd.
Step three: test (c). \( h(-x) = (-x)^2 + (-x) = x^2 - x \). This is neither \( x^2 + x \) nor \( -x^2 - x \).
Step four: confirm with numbers. Since an example can disprove, use one: \( h(2) = 6 \) and \( h(-2) = 4 - 2 = 2 \). For even, we would need \( h(-2) = 6 \); for odd, \( h(-2) = -6 \). Neither holds, so \( h \) is neither even nor odd.
Step five: test (d). \( k(-x) = \dfrac{-x}{(-x)^2 + 1} = \dfrac{-x}{x^2 + 1} = -k(x) \). So \( k \) is odd. The denominator is even, since it contains only \( x^2 \), and dividing an odd numerator by an even denominator gives an odd function.
Step six: state the product rules. Even times even is even, odd times odd is even, and even times odd is odd.
Step seven: prove one rule. If \( f \) and \( g \) are both odd, let \( p(x) = f(x)g(x) \). Then \( p(-x) = f(-x)g(-x) = \big(-f(x)\big)\big(-g(x)\big) = f(x)g(x) = p(x) \). The two minus signs cancel, so the product is even.
Step eight: connect it to what is coming. These rules explain why \( \sin x \) is odd and \( \cos x \) is even in Unit 4: the coordinates of a point on the unit circle reflect that way when the angle changes sign. They also mean that \( \sin x \cos x \) is odd and \( \cos^2 x \) is even, which can be read off without any computation.
Problems 1 to 5 are routine. Problems 6 to 10 are multi-step or ask for reasoning.
- Is \( f(x) = x^2 \) even, odd or neither?
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\( f(-x) = (-x)^2 = x^2 = f(x) \). Even
- Is \( f(x) = x^5 - x \) even, odd or neither?
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\( f(-x) = -x^5 + x = -(x^5 - x) = -f(x) \). Odd
- Is \( f(x) = x^2 + 2x \) even, odd or neither?
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\( f(-x) = x^2 - 2x \), which is neither \( f(x) \) nor \( -f(x) \). Neither
- The point \( (3, 7) \) lies on the graph of an even function. Name another point.
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An even function has \( f(-3) = f(3) = 7 \). \( (-3, 7) \)
- The point \( (3, 7) \) lies on the graph of an odd function. Name another point.
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An odd function has \( f(-3) = -f(3) = -7 \). \( (-3, -7) \)
- Classify \( f(x) = |x| - x^2 \).
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\( f(-x) = |-x| - (-x)^2 = |x| - x^2 = f(x) \). Even
- Classify \( f(x) = x^3 - x^2 \), and explain the numerical check.
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\( f(-x) = -x^3 - x^2 \). This is not \( f(x) \), and \( -f(x) = -x^3 + x^2 \) is not it either. Numerical check: \( f(1) = 0 \) and \( f(-1) = -1 - 1 = -2 \). For even we would need \( f(-1) = 0 \), and for odd we would need \( f(-1) = -0 = 0 \). Neither holds. Neither
- Classify \( g(x) = \dfrac{x^3}{x^2 + 4} \).
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\( g(-x) = \dfrac{(-x)^3}{(-x)^2 + 4} = \dfrac{-x^3}{x^2 + 4} = -g(x) \). Odd
- Prove that the sum of two odd functions is odd.
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Let \( f \) and \( g \) be odd and \( s(x) = f(x) + g(x) \). \( s(-x) = f(-x) + g(-x) = -f(x) - g(x) = -\big(f(x) + g(x)\big) = -s(x) \). So \( s \) is odd
- Show that the only function that is both even and odd is the zero function, and explain
what this says about an odd function at 0.
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If \( f \) is even, \( f(-x) = f(x) \). If it is also odd, \( f(-x) = -f(x) \). Equating: \( f(x) = -f(x) \), so \( 2f(x) = 0 \) and \( f(x) = 0 \) for every \( x \). For an odd function alone, put \( x = 0 \): \( f(0) = -f(0) \), so \( f(0) = 0 \), provided 0 is in the domain. This is why \( f(x) = x + 1 \) cannot be odd: \( f(0) = 1 \). Only \( f(x) = 0 \); an odd function must pass through the origin if defined there