I have been stuck at this question: I have $$f(x)=\cos(\pi \cdot x)$$$$g(x)=\frac{7\cdot x}{6}$$ and $$h(x)=f(g(x))$$
and i am asked to compute the range for $h(x)$.
My solution: $$h(x)=\cos(\pi \cdot \frac{7x}{6})$$ so the highest value the function can output is $1$ and the lowest is $-1$. My instructor says it's not the right answer. How do i go about finding the range of this function? Thank you in advance.
EDIT
$f:\mathbb{Q}\to\mathbb R$
$g:\mathbb{Z}\to\mathbb{Q}$
Your composite function then is
$$f\circ g:\mathbb{Z}\rightarrow \mathbb{R}$$
But now, by inputting only multiples of $\frac{7\pi}{6}$ into the cosine function, and because cosine is a periodic function, you are only going to get specific values for your range. These are your "special" angle values because you will never have an angle as an input to cosine that is NOT a multiple of $\frac{\pi}{6}$. So your range will be $$\left\{-1,-\frac{\sqrt{3}}{2},-\frac{1}{2},0,\frac{1}{2},\frac{\sqrt{3}}{2},1\right\}$$
Since $f(x) = \cos(\pi x)$ has periodicity $2$, we have $h(x) = f(g(x)) = f(\frac{7x}{6}+2m)$ with $m$ integer. Now note that $h(x)=h(x+12)$.
Thus your codomain will be determined by $x\in\{0,1,\dots,11\}$. The range is then determined by the unique subset of values from $\{h(0),h(1),\dots,h(11)\}$.
Normally range is taken to mean the image (set of possible outputs). Perhaps I'm wrong but you seem to want to know the global minimum and global maximum of $h$. The global minimum of $h$ is $-1$ and the global maximum is $1$.
$h(12) = \mathrm{cos}(14 \pi) = 1$, and $h(6) = \mathrm{cos}(7 \pi) = -1$.
Since $\forall x \in \mathbb{R}$ $\mathrm{cos}(x) \in [-1,1]$, we have $\forall x \in \mathbb{Q}$ $f(x) \in [-1,1]$. This means that $h$ does not take values above $1$ or below $-1$.
