7.4 Symmetries

This function is symmetric across the $y$-axis.

even

This function is symmetric across the $y$-axis, so it is even.

There are two.

odd, periodic

The function is not even, because it is not mirrored across the $y$-axis.

Assuming it continues as shown, the function is periodic, because the unit shown below is repeated:

Additionally, $f(x)$ is odd. In a function with odd symmetry, if we mirror the right-hand portion of the curve (the portion to the right of the $y$-axis) across both the $y$-axis and the $x$-axis, it lines up with the left-hand portion of the curve.

Since reflecting the right-hand portion of the graph across the $y$-axis, then the $x$-axis, gives us $f(x)$, we conclude $f(x)$ is odd.

Since the function is even, you only have to reflect the portion shown across the $y$-axis to complete the sketch.

Since the function is even, we simply reflect the portion shown across the $y$-axis to complete the sketch.

Since the function is odd, to complete the sketch, reflect the portion shown across the $y$-axis, then the $x$-axis.

Since the function is odd, to complete the sketch, we reflect the portion shown across the $y$-axis (shown dashed), then the $x$-axis (shown in red).

A function is even if $f(-x)=f(x)$.

A function is even if $f(-x)=f(x)$.

So, $f(x)$ is even.

A function is even if $f(-x)=f(x)$.

So, $f(x)$ is even.

Its period is not $2\pi$.

For any real number $x$, we will show that $f(x)=f(x+4\pi)$.

So, $f(x)$ is periodic.

For any real number $x$, we will show that $f(x)=f(x+4\pi)$.

So, $f(x)$ is periodic.

Simplify $f(-x)$ to see whether it is the same as $f(x)$, $-f(x)$, or neither.

even

$f(x)$ is not periodic. (You don't really have to justify this, but if you wanted to, you could say something like this. Notice $f(0)=1$. Whenever $x>10$, $f(x)>1$. Then the value of $f(0)$ is not repeated indefinitely, so $f(x)$ is not periodic.)

To decide whether $f(x)$ is even, odd, or neither, simplify $f(-x)$:

Since $f(-x)=f(x)$, our function is even.

Simplify $f(-x)$ to see whether it is the same as $f(x)$, $-f(x)$, or neither.

none

It should be clear that $f(x)$ is not periodic. (If you wanted to justify this, you could note that $f(x)=0$ has exactly two solutions, $x=0,\,-{5}$. Since the value of $f(0)$ is repeated only twice, and not indefinitely, $f(x)$ is not periodic.)

To decide whether $f(x)$ is odd, even, or neither, we simplify $f(-x)$.

We see that $f(-x)$ is not equal to $f(x)$ or to $-f(x)$. For instance, when $x=1$:

Since $f(-x)$ is not equal to $f(x)$ or to $-f(x)$, $f(x)$ is neither even nor odd.

Find the smallest value $k$ such that $f(x+k)=f(x)$ for any $x$ in the domain of $f$.

You may use the fact that the period of $g(X)=\tan X$ is $\pi$.

1

Recall the period of $g(X)=\tan X$ is $\pi$.

Replacing $X$ with $\pi x$:

The period of $f(x)$ is 1.

It is true that $f(x)=f(x+2\pi)$ for every $x$ in the domain of $f(x)$, but the period is not $2\pi$.

$\pi$

Let's consider $g(x)=\tan(3x)$ and $h(x)=\sin(4x)$ separately. Recall that $\pi$ is the period of tangent.

Replacing $X$ with $3x$:

So, the period of $g(x)=\tan(3x)$ is $\dfrac{\pi}{3}$.

Similarly, $2\pi$ is the period of sine.

Replacing $X$ with $4x$:

So, the period of $h(x)=\sin(4x)$ is $\dfrac{\pi}{2}$.

All together, $f(x)=g(x)+h(x)$ will repeat when both $g(x)$ and $h(x)$ repeat. The least common integer multiple of $\dfrac{\pi}{3}$ and $\dfrac{\pi}{2}$ is $\pi$. Since $g(x)$ repeats every $\dfrac{\pi}{3}$ units, and $h(x)$ repeats every $\dfrac{\pi}{2}$ units, they will not both repeat until we move $\pi$ units. So, the period of $f(x)$ is $\pi$.