Even and odd functions, can someone help me understand this?

Question

Functions $f$, $g$, and $h$ are defined as follows: \begin{align} f&: \mathbb{R} \to \mathbb{R}, &f(x) &= x^2 - 10x + 9 \\ g&: \mathbb{R} \to \mathbb{R}, &g(x) &= x^4 - 10x^2 + 9 \\ f&: \mathbb{R} \to \mathbb{R}, &f(x) &= x^3 - 4x \end{align}

For each of these functions:

a) Determine intersection points between the graph of the function and coordinate axes.

b) Sketch the graph of the function. Is the graph symmetric about some straight line or at some point?

c) Determine whether the function is even, odd, or neither even nor odd.

a) For $f$, we have $f(x) = x^2 - 10x + 9$. By setting $f(x) = 0$ and solving for $x$ we get $x = 1$ and $x = 9$. So the intersection points are $(1, 0)$ and $(9, 0)$?

For $g$, we have $g(x) = x^4 − 10x^2 + 9$. For $h$, we have $h(x) = x^3 − 4x$.

I know we've to set them equal to $0$ and solve for $x$ to find intersection points, but I have trouble completing the square and would need your help.

Not there yet with b).

I know that odd means $f(-x) = -f(x)$, and even means $f(-x) = f(x)$, for all $x$. Not quite sure what it means for it to be neither.

Answer 1

$g$ is $f(x^2)$, so the intersection points of $g$ with the $x$ axis are the square roots of the intersection points of $f$ with the $x$ axis.

For $h$ it's clear that $h(0)=0$ so you have one intersection point. Then you can write $h(x)=x(x^2-4)$ and you might be able to guess the other two intersection points.

Most functions aren't either odd or even. For example $\sin(x)$ is odd, and $\cos(x)$ is even, but any sum of $\sin$ and $\cos$ is neither even nor odd.

Answer 2

This answer addresses the symmetry properties of certain graphs and the notions of even and odd functions. First of all, the property of the graph of a function having a symmetry (either mirror reflection or rotation/origin reflection) is rare. Most functions have neither.

Even function. Geometrically speaking, the graph of an even function has mirror symmetry about the $y$-axis, i.e. $(x, y)$ is on the graph if and only if $(-x, y)$ is on the graph. But that means that $f(-x) = y = f(x)$, which is the algebraic condition you mention at the bottom.

The graph of $f$ that you're working with is a parabola (since it's a quadratic polynomial function), and you probably know that parabolas have a reflection symmetry, but in this case, the axis of symmetry is the vertical line $x = 5$ (the average of the roots). As we can work out, this means that $f(5-x) = f(5+x)$ for all $x$, but that symmetry doesn't have a name like even. You can even see this by completing the square in $x$. (Try it yourself before revealing the spoiler.

$$f(x) = x^2 - 10x + 9 = x^2 - 10x + 25 - 16 = (x - 5)^2 - 16 $$

You can shift to a new horizontal coordinate, say $X = x - 5$, so $x = X + 5$ and work out what $f(X)$ is:

$$f(X) = f(x - 5) = (x - 5)^2 - 16 = X^2 - 16, $$

and in this form, it's easy to see that $f(X) = f(-X)$ for all $X$, it has symmetry about the vertical line $X = 0$. Translating this back into the original horizontal coordinate $x$, we see that $f(5-x) = f(5+x)$ and that the vertical line $x = 5$ is the axis of symmetry of the parabola.

You can probably work out that your function $g$ is in fact even. (It turns out that for polynomial functions, being even means having only nonzero terms of even degree, and the quartic polynomial $g$ has terms of degrees $4$, $2$, and $0$.) This general observation about polynomials also demonstrates why $f$ is not even since it has a nonzero linear term, i.e. a term of odd degree.

Odd function. Geometrically speaking, the graph of an odd function has mirror symmetry about the origin, i.e. $(x, y)$ is on the graph if and only if $(-x, -y)$ is on the graph. But that means that $f(-x) = -y = -f(x)$, which is the algebraic condition you mention at the bottom. By the way, this is equivalent to the symmetry whereby we take the graph and rotate it half-way around $(180^\circ)$ about the origin, but that's not necessary here.

You can verify that your function $h$ satisfies this property by evaluating at an arbitrary $-x$ expression and simplifying it to match the value at $x$:

$$h(-x) = (-x)^3 - 4(-x) = -x^3 + 4x = -(x^3 - 4x) = -h(x) $$

One property that all such odd functions follows immediately from the fact that $-0 = 0$ so $h(0) = h(-0) = -h(0)$. But what number equals its opposite? Only zero. Hence, $h(0) = 0$, i.e. assuming that the function is defined at $0$, the graph passes through the origin, the center of the $180^\circ$ rotation.