To show that $f$ is 1-1, you could show that $$f(x)=f(y)\Longrightarrow x=y.$$
So, for example, for $f(x)={x-3\over x+2}$:
Suppose ${x-3\over x+2}= {y-3\over y+2}$. Then:
\begin{align*}
&{x-3\over x+2}= {y-3\over y+2} \\
\Longrightarrow& (y+2)(x-3)= (y-3)(x+2)\\
\iff& yx+2x-3y-6= yx-3x+2y-6\\
\iff&2x-3y =-3x+2y\\
\iff&2x+3x =2y+3y\\
\iff&5x =5y\\
\iff&x=y
\end{align*}
So $f(x)={x-3\over x+2}$ is 1-1.
I'll leave showing that $f(x)={{x-3}\over 3}$ is 1-1 for you.
Alternatively, to show that $f$ is 1-1, you could show that $$x\ne y\Longrightarrow f(x)\ne f(y).$$
Or, for a differentiable $f$ whose derivative is either always positive or always negative, you can conclude $f$ is 1-1 (you could also conclude that $f$ is 1-1 for certain functions whose derivatives do have zeros; you'd have to insure that the derivative never switches sign and that $f$ is constant on no interval).
You would discover that a function $g$ is not 1-1, if, when using the first method above, you find that the equation is satisfied for some $x\ne y$. For example, take $g(x)=1-x^2$. Then
$$
\eqalign{
&g(x)=g(y)\cr
\iff&{1-x^2}= {1-y^2} \cr
\iff&-x^2= -y^2\cr
\iff&x^2=y^2\cr}
$$
The above equation has $x=1$, $y=-1$ as a solution. So, there is $x\ne y$ with $g(x)=g(y)$; thus $g(x)=1-x^2$ is not 1-1.
Of course, to show $g$ is not 1-1, you need only find two distinct values of the input value $x$ that give $g$ the same output value.
Although you rightfully point out that the graphical method is unreliable; it is still instructive to consider the methods used and why they work:
Graphically, you can use either of the following:
- Use the "Horizontal Line Test":
$f$ is 1-1 if and only if every horizontal line intersects the graph
of $f$ in at most one point. Note that this is just the graphical
interpretation of "if $x\ne y$ then $f(x)\ne f(y)$"; since the
intersection points of a horizontal line with the graph of $f$ give
$x$ values for which $f(x)$ has the same value (namely the $y$-intercept of the line).
- Use the fact that a continuous $f$ is 1-1 if and only if $f$ is either
strictly increasing or strictly decreasing. This, of course, is the case if $f$ is differentiable and the derivative is always positive or always negative (with perhaps being zero at "isolated" points).
(Note this method applies to only the green function below.)
Amplitude, period and phase shift of can be recovered from the graph by noticing the coordinates of the peaks and troughs of the wave.
Let $y_{peak}$ and $y_{trough}$ denote the $y$-coordinates of the peaks and troughs of the wave. Then for the amplitude we have
$$ A=\frac{1}{2}\left(y_{peak}-y_{trough}\right) $$
From your graph there is no indication of the vertical scale, but let us suppose that the horizontal dashed lines are one unit apart. Then we would have $y_{peak}=2$ and $y_{trough}=-6$. This would give $A=\frac{1}{2}(2-(-6))=4$.
We can also use $y_{peak}$ and $y_{trough}$ to find the vertical shift $D$.
$$ D=\frac{1}{2}\left(y_{peak}+y_{trough}\right) $$
So, for your example, $D=\frac{1}{2}(2+(-6))=-2$.
This leaves the values of $B$ and $C$. But first, we must find the period $P$, which is straightforward.
The period P is the horizontal distance between two successive peaks of the graph. For your graph this would be a distance $P=3\pi$.
The value of $B$ is then found from
$$ B=\frac{2\pi}{P} $$
For your graph, then, we have $B=\frac{2\pi}{3\pi}=\frac{2}{3}$.
Finally we have the phase shift $\phi$.
The phase shift for the sine and cosine are computed differently. The easiest to determine from the graph is the phase shift of the cosine.
Let $x_{peak}$ denote the $x$-coordinate of the peak closest to the vertical axis. For your graph, we have $x_{peak}=0$.
$$\phi=x_{peak}\text{ for the cosine graph}$$
$$\phi=\left(x_{peak}-\frac{P}{4}\right)\text{ for the sine graph}$$
For your graph this gives phase shift $\phi=0$ for the cosine graph and phase shift $\phi=0-\frac{3\pi}{4}=-\frac{3\pi}{4}$ for the sine graph.
But for your equations, we need the value of $C$. The value of $C$ is found from the values of $\phi$ and $B$.
The equation for $C$ in terms of the phase shift $\phi$ is
$$ C=B\phi $$
So if we use the cosine function to model your graph we have $C=0$ and for the sine graph we have $C=\frac{2}{3}\cdot\left(-\frac{3\pi}{4}\right)=-\frac{\pi}{2}$.
So we have for both sine and cosine
- $A=4$
- $D=-2$
- $B=\frac{2}{3}$
For cosine, $C=0$ and for sine, $C=-\frac{\pi}{2}$.
Thus your graph can be represented by either of the two equations
$$ y=4\cos\left(\frac{2}{3}x \right)-2 $$
$$ y=4\sin\left(\frac{2}{3}x+\frac{\pi}{2}\right)-2 $$
Best Answer
The correct method is to graph $y=\frac{1}{2}\cos(x)$ first, then follow the same procedure as you did for graphing $y=\sec x$. Why so? Because you want a pivot point about which you're flipping the graph of $\cos(x)$ and for that both functions must have the same magnitude, which they do at $x=n\pi, n\in\mathbb{Z}$.