How can we use the derivative of f(x) to determine if its injective? I'm not sure if I have the right idea: suppose you find the derivative of a function and $f'(x) > 0 $ or $f'(x)< 0$, does that mean it is injective? 
The derivative has to be one or the other to be injective? Thank you!
 A: If $f$ is defined on an open interval, and $f'(x) > 0$ at all points on its domain, then $f$ is strictly increasing, and hence injective. Similarly, if $f'(x) < 0$ for all $x$ instead, then it is strictly decreasing, and still injective.
The proof of this is the Mean Value Theorem. If $a < b$, then the MVT says there exists some $c \in (a, b)$ such that
$$f'(c) = \frac{f(b) - f(a)}{b - a}.$$
In particular, if we know that $f'(x) > 0$ for all $x$, then $f'(c) > 0$, hence $f(b) - f(a) > 0$. So, no distinct points $a \neq b$ can have the same function value.
Is there a converse to this result? Sort of yes, but it's a little more complex than your guess. The Intermediate Value Theorem guarantees that a continuous, injective real function, defined on an interval, is either strictly increasing or decreasing, meaning that $x < y \implies f(x) < f(y)$ for all $x, y$ or respectively, $x < y \implies f(y) < f(x)$ for all $x, y$. However, this doesn't guarantee $f'(x) > 0$ for all $x$ (respectively $f'(x) < 0$ for all $x$), even if the function is differentiable. For example,
$$f(x) = x^3$$
is a strictly increasing function, but $f'(0) = 0$. Instead, all we can say is $f'(x) \ge 0$ for all $x$ and there is no non-trivial interval on which $f'$ is constantly $0$. It's OK for $f'(x) = 0$ for some specific $x$, so long as there are points nearby where the derivative is strictly positive.
Also, beware that the condition that $f$ is defined on an interval is vitally important. For example, the function $f(x) = \frac{1}{x}$ on its natural domain $\Bbb{R} \setminus \{0\}$ has an always strictly negative derivative $f'(x) = -\frac{1}{x^2}$, but is not a decreasing function. In particular, $-1 < 1$, but $f(-1) < f(1)$.
