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Section4.6Even Functions and Odd Functions

An even function, \(f\text{,}\) is a function with the property that \(f(-x)=f(x)\) for every value of \(x\) in the domain of \(f\text{.}\)

An odd function, \(f\text{,}\) is a function with the property that \(f(-x)=-f(x)\) for every value of \(x\) in the domain of \(f\text{.}\)

Example4.6.1

Determine whether the function \(f(x)=x^4-3x^2+7\) is an even function, and odd function, or neither an even function nor an odd function.

Solution

We need to simplify the formula for \(f(-x)\text{.}\) If it is identical to the formula for \(f(x)\text{,}\) we can declare that \(f\) is an even function. If not, we need to simplify the formula for \(-f(x)\) and compare that to the simplified formula for \(f(-x)\text{.}\) If the two formulas are the same, we can state that \(f\) is an odd function. If those last two formulas are not identical, we state that \(f\) is neither an even function nor an odd function.

\begin{align*} f(-x)\amp=(-x)^4+(-x)^2+7\\ \amp=x^4+x^2+7 \end{align*}

We compare the formula we just simplified to the stated formula for \(f(x)\) and observe that they are identical. Because \(f(-x)\) and \(f(x)\) have the same formula, \(f\) is an even function.

Example4.6.2

Determine whether the function \(f(x)=-x^5+14x^3\) is an even function, and odd function, or neither an even function nor an odd function.

Solution

We begin by simplifying \(f(-x)\text{.}\)

\begin{align*} f(-x)\amp=-(-x)^5+14(-x)^3\\ \amp=x^5-14x^3 \end{align*}

This formula is not the same as the formula for \(f(x)\text{,}\) so we now state and simplify the formula for \(-f(x)\text{.}\)

\begin{align*} -f(x)\amp=-\left(-x^5+14x^3\right)\\ \amp=x^5-14x^3 \end{align*}

This formula is identical to the simplified formula for \(f(-x)\text{,}\) so we conclude that \(f\) is an odd function.

Example4.6.3

Determine whether the function \(f(x)=-3x^9-31x^6+16x\) is an even function, and odd function, or neither an even function nor an odd function.

Solution

We begin by simplifying the formula for \(f(-x)\text{.}\)

\begin{align*} f(-x)\amp=-3(-x)^9-31(-x)^6+16(-x)\\ \amp=3x^9-31x^6-16x \end{align*}

We observe that this formula is not the same as the formula for \(f(x)\text{,}\) so we proceed to simplify \(-f(x)\text{.}\)

\begin{align*} -f(x)\amp=-\left(-3x^9-31x^6+16x\right)\\ \amp=3x^9+31x^6-16x \end{align*}

We observe that the formula for \(f(-x)\) is also not the same as the formula for \(f(-x)\) (the signs of the coefficients on the second term are not the same), so we can state with confidence that \(f\) is neither an even function nor an odd function.

You may or may have no noticed that in the first example all of the terms had even degree (the power on \(x\)) and the function turned out to be even. Similarly, in the second example all of the terms had odd degree and function turned out to be odd. Finally, in the third example there were both terms with even degree and terms with odd degree, and the function turned out to be neither even nor odd. That might seem like a pretty big coincidence, but I suspect this phenomenon is the principle impetus for the terms even function and odd function, because for polynomial functions, the degrees of the terms always correspond the function type (odd, even, or neither).

The correspondence between term degree and function type only applies to polynomial functions, as we will see in the next example.

Example4.6.4

Determine whether the function \(f(x)=\frac{7x^{11}}{4x^5-6x}\) is an even function, and odd function, or neither an even function nor an odd function.

Solution

We begin by simplifying the formula for \(f(-x)\text{.}\)

\begin{align*} f(-x)\amp=\frac{7(-x)^{11}}{4(-x)^5-6(-x)}\\ \amp=\frac{-7x^{11}}{-4x^5+6x}\\ \amp=\frac{-1 \cdot 7x^{11}}{-1 \cdot \left(4x^5-6x\right)}\\ \amp=\frac{7x^{11}}{4x^5-6x} \end{align*}

We see that the formula for \(f(-x)\) is identical to the formula for \(f(x)\) and conclude that \(f\) is an even function. Let's note that \(f\) tuned out to be an even function even though every term has odd degree. This does not contradict what was stated above, however, because the function we were analyzing was not a polynomial function.

The graphs of even functions and odd functions each have there own unique graphical properties.

A function, \(y=f(x)\text{,}\) is an even function if and only if its graph is symmetric across the \(y\)-axis. This is because for every point on the graph, \((a,b)\text{,}\)the point \((-a,b)\) also lies on the graph. So if we fold the plot across the \(y\)-axis, every point on the function that lies to the left of the axis will lie atop its twin that lies to the right of the \(y\)axis.

The function shown in FigureĀ 4.6.5 is symmetric across the \(y\)-axis and therefore we know that it is an even function,

A function resembling a curved capital M that is symmetric across the \(y\)-axis.  Three points on the graph are \((-3,2)\text{,}\) \((3,2)\) and \((0,2)\text{.}\)
Figure4.6.5The graph of an even function

A function, \(y=f(x)\text{,}\) is an odd function if and only if its graph is symmetric about the origin. Symmetry about the origin is a rotational symmetry, which can be a little more difficult to discern than symmetry across the \(y\)-axis. Image that you had two copies of the same graph showing a function that is symmetric about the origin, let's call them copy A and copy B. If we rotated copy A \(180^{\circ}\) and laid it atop copy B, the the two curves would lie atop one another - that's what is meant by symmetry about the origin.

The function shown in FigureĀ 4.6.6 is symmetric about the origin and therefore we know that it is an odd function.

The graph of a function that is symmetric about the origin.  The function includes the point \((1,1)\) and the point \((1,-1)\text{.}\)  To the left of the \(y\)-axis there is a curve that lies entirely in Quadrant III.  The curve approaches both the \(x\)-axis and the \(y\)-axis in an asymptotic manner.  To the right of the \(y\)-axis there is a curve that lies entirely in Quadrant I.  The curve approaches both the \(x\)-axis and the \(y\)-axis in an asymptotic manner.
Figure4.6.6The graph of an odd function

Subsection4.6.1Exercises

Determine whether each given function is an odd function, an even function, or neither an even function nor an odd function.

1

\(f(x)=\abs{x}-7x^8\)

Solution

We begin by simplifying the expression \(f(-x)\text{.}\)

\begin{align*} f(x)\amp=\abs{-x}-7(-x)^8\\ \amp=\abs{x}-7x^8 \end{align*}

The formula for \(f(-x)\) is identical to the formula for \(f(x)\text{,}\) so \(f\) is an even function.

2

\(f(x)=12x^7-\abs{x}\)

Solution

We begin by simplifying the expression \(f(-x)\text{.}\)

\begin{align*} f(-x)\amp=12(-x)^7-\abs{-x}\\ \amp=-12x^7-\abs{x} \end{align*}

The formula for \(f(-x)\) is not the same as the formula for \(f(x)\text{,}\) so the function is not even. Let's check if the function is odd by first finding the formula for \(-f(x)\text{.}\)

\begin{align*} -f(x)\amp=-\left(12x^7-\abs{x}\right)\\ \amp=-12x^7+\abs{x} \end{align*}

The formula for \(f(-x)\) is also not the same as the formula for \(-f(x)\text{,}\) so we conclude that \(f\) is neither an even function nor an odd function.

3

\(f(x)=\frac{x}{3x^3-5x}\)

Solution

We begin by simplifying the expression \(f(-x)\text{.}\)

\begin{align*} f(-x)\amp=\frac{-x}{3(-x)^3-5(-x)}\\ \amp=\frac{-x}{-3x^3+5x}\\ \amp=\frac{-1 \cdot x}{-1 \cdot \left(3x^3-5x\right)}\\ \amp=\frac{x}{3x^3-5x} \end{align*}

The formula for \(f(-x)\) is identical to the formula for \(f(x)\text{,}\) so \(f\) is an even function.

4

\(f(x)=\frac{1}{3x^3-5x}\)

Solution

We begin by simplifying the expression \(f(-x)\text{.}\)

\begin{align*} f(-x)\amp=\frac{1}{3(-x)^3-5(-x)}\\ \amp=\frac{1}{-3x^3+5x} \end{align*}

The formula for \(f(-x)\) is not the same as the formula for \(f(x)\text{,}\) so the function is not even. Let's check if the function is odd by first finding the formula for \(-f(x)\text{.}\)

\begin{align*} -f(x)\amp=-\frac{1}{3x^3-5x}\\ \amp=\frac{1}{-1 \cdot \left(3x^3-5x\right)}\\ \amp=\frac{1}{-3x^3+5x} \end{align*}

The formula for \(f(-x)\) is identical to the formula for \(-f(x)\text{,}\) so \(f\) is an odd function.

5

\(f(x)=\frac{1}{3x^3-5x^2}\)

Solution

We begin by simplifying the expression \(f(-x)\text{.}\)

\begin{align*} f(-x)\amp=\frac{1}{3(-x)^3-5(-x)^2}\\ \amp=\frac{1}{-3x^3-5x^2} \end{align*}

The formula for \(f(-x)\) is not the same as the formula for \(f(x)\text{,}\) so the function is not even. Let's check of the function is odd by first finding the formula for \(-f(x)\text{.}\)

\begin{align*} -f(x)\amp=-\frac{1}{3x^3-5x^2}\\ \amp=\frac{1}{-1 \cdot \left(3x^3-5x^2\right)}\\ \amp=\frac{1}{-3x^3+5x} \end{align*}

The formula for \(f(-x)\) is also not the same as the formula for \(-f(x)\text{,}\) so we conclude that \(f\) is neither an even function nor an odd function.

6

\(-5x^2+\frac{7}{x^4} \)

Solution

We begin by simplifying the expression \(f(-x)\text{.}\)

\begin{align*} f(-x)\amp=-5(-x)^2+\frac{7}{(-x)^4}\\ \amp=-5x^2+\frac{7}{x^4} \end{align*}

The formula for \(f(-x)\) is identical to the formula for \(f(x)\text{,}\) so \(f\) is an even function.

7

The function shown in FigureĀ 4.6.7.

The graph of a function that is symmetric across the \(y\)-axis.  The function consists of two distinct pieces.  One piece is on the left side of the \(y\)-axis and consists of a nearly vertical section that points downward close to the \(y\)-axis and turns to the left, immediately becoming almost horizontal.  The other piece is on the right side of the \(y\)-axis and consists of a nearly vertical section that points downward close to the \(y\)-axis and turns to the right, immediately becoming almost horizontal.
Figure4.6.7Odd, Even, or Neither?
Solution

The function is symmetric across the \(y\)-axis, so it is an even function.

8

The function shown in FigureĀ 4.6.8.

The graph of a function that is symmetric about the origin.. Three points on the function are \((-2,0)\text{,}\) \((0,0)\text{,}\) and \((2,0)\text{.}\)  Over the interval \((-2,0)\) the function lies close to the \(x\)-axis in Quadrant II.  Over the interval \((0,2)\) the curve lies close to the \(x\)-axis in Quadrant IV.  Over the interval \((-\infty,-2)\) the function points almost straight down in Quadrant III with an arrow at the end.  Over the interval \((2,\infty)\) the function points almost straight up in Quadrant I with an arrow at the end.
Figure4.6.8Odd, Even, or Neither?
Solution

The function is symmetric about the origin, so it is an odd function.

9

The function shown in FigureĀ 4.6.9.

The graph of a function that is crazy squiggly yet never-the-less symmetric across the \(y\)-axis.  Three points on the function are \((-5,0)\text{,}\) \((0,-5)\text{,}\) and \((5,0)\text{.}\)
Figure4.6.9Odd, Even, or Neither?
Solution

The function is symmetric across the \(y\)-axis, so it is an even function.

10

The function shown in FigureĀ 4.6.10.

The graph of a function that is symmetric across the line \(y=2\text{.}\)  The function lies entirely in Quadrants I and II.  The line \(y=2\) acts as a vertical asymptote and the \(x\)-axis acts as a horizontal asymptote.
Figure4.6.10Odd, Even, or Neither?
Solution

While the function does have vertical line symmetry, the function is not specifically symmetric across the \(y\)-axis, so it is not an even function. The function is also clearly not symmetric about the origin, so it is also not an odd function. In summary, the function is neither even nor odd.

11

The function shown in FigureĀ 4.6.11.

The graph of a squiggly function that has no sort of symmetry.
Figure4.6.11Odd, Even, or Neither?
Solution

The function is neither symmetric across the \(y\)-axis nor symmetric about the origin, so it is neither even nor odd.