Help and verification on $1$-$1$ functions and their images $a) \space f(x) = 3x-7$ $E=\mathbb{R}$
The image of $f$ on $E$ is $\mathbb{R}$ correct?
$b) f(x) = e^{(\frac{1}{x})}$ with domain $(0, \infty)$
The image of $f$ on $E$ is $(0,\infty)$ correct?
c) I can verify that the following function is $1-1$, but I cannot find the image of the following function:
$f(x) = \tan(x) \space E= (\frac{\pi}{2},\frac{3\pi}{2})$
d) $f(x) = x^2 + 2x - 5 \space E = (-\infty,-6]$
$\frac{dy}{dx} = 2x + 2$
$f'(x) \lt 0$ on $(-\infty, -6]$ thus $f$ is decreasing on the interval. So $f$ is $1 - 1$
The image of $f$ is $[19, \infty)$ 
Is my proof of $1-1$ correct? what about the image?
$e) f(x) = \frac{x}{x^2+1} \space E = [-1,1]$
$\frac{dy}{dx} \space = \space \frac{x^2 + 1 - 2x^2}{(x^2+1)^2}$
$ = \frac{1-x^2}{(x^2+1)^2}$
The image of $f$ is $-\frac{1}{2},\frac{1}{2}$ is any of this correct? H0w do I show its $1 - 1$? Plugging in the domain gives 0.
f) $f(x) = 3x - \vert x \vert + \vert x - 2 \vert E = \mathbb{R}$
How do I show the image and prove that it is $1-1$? 
 A: 
$f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3x - 7$

You are correct that the range is $\mathbb{R}$.
You can show the function is injective (one-to-one) by showing that if $f(x_1) = f(x_2)$, then $x_1 = x_2$.  Alternatively, you can show that the function is strictly increasing.

$f: (0, \infty) \to \mathbb{R}$ defined by $f(x) = e^{1/x}$

As Gerry Myerson indicated in the comments, your range is wrong.
\begin{align*}
\lim_{x \to \infty} f(x) & = \lim_{x \to \infty} e^{1/x} = \lim_{t \to 0^+} e^t = 1\\
\lim_{x \to 0^+} f(x) & = \lim_{x \to 0^+} e^{1/x} = \infty
\end{align*}
Since $f$ is continuous on $(0, \infty)$, the range of the function is $(1, \infty)$.
You can show the function is injective by demonstrating that it is strictly decreasing on its domain.

$f: \left(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right) \to \mathbb{R}$ defined by $f(x) = \tan x$

You are correct that $f$ is injective.
As for the range, one way to define the tangent function is as the point where the line containing the terminal side of the angle intersects the line $x = 1$.

As $x$ increases from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$, $f(x)$ increases from $-\infty$ to $\infty$.  Since $f$ is periodic with period $\pi$, it has the same range in the interval $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$. The graph of the tangent function on the domain 
$$\bigcup_{k \in \mathbb{Z}} \left(-\frac{\pi}{2} + k\pi, \frac{\pi}{2} + k\pi\right)$$
is shown below.


$f: (-\infty, -6] \to \mathbb{R}$ defined by $f(x) = x^2 + 2x - 5$

Your solution is correct.

$f: [-1, 1] \to \mathbb{R}$ defined by $f(x) = \dfrac{x}{x^2 + 1}$

You correctly calculated the derivative.  Since the function is differentiable, it is continuous.  Moreover, notice that 
$$f'(x) = \frac{1 - x^2}{(x^2 + 1)^2} > 0$$
in the open interval $(-1, 1)$.  Hence, it is strictly increasing on the closed interval $[-1, 1]$.  Therefore, the function is injective.  Since $f(-1) = -1/2$ and $f(1) = 1/2$, the function has range $[-1/2, 1/2]$.

$f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 3x - |x| + |x - 2|$

Observe that $|x| = x$ when $x \geq 0$ and that $|x| = -x$ when $x < 0$.  Also, observe that $|x - 2| = x - 2$ when $x \geq 2$ and that $|x - 2| = -x + 2$ when $x < 2$.  Hence, 
\begin{align*}
f(x) & = \begin{cases}
3x + x - x + 2 & \text{if $x < 0$}\\
3x - x - x + 2 & \text{if $0 \leq x < 2$}\\
3x - x + x - 2 & \text{if $x \geq 2$}
\end{cases}\\
& = \begin{cases}
3x + 2 & \text{if $x < 0$}\\
x + 2 & \text{if $0 \leq x < 2$}\\
3x - 2 & \text{if $x \geq 2$}
\end{cases}
\end{align*}
To show that the function is injective, show that it is strictly increasing. As for finding the image, try graphing the piecewise function.
