What possible values of $a$ lead to the inequality $|f(x_1) - f(x_2)| \gt |g(x_1) - g(x_2)|$. Let $a \in \mathbb{R}$ and define
$$f(x) = e^x, \quad g(x) = x^2 + ax + 1.$$
Find the possible values of $a$ such that for all $x_1, x_2 \in [0, 2], x_1 \not = x_2$, $$|f(x_1) - f(x_2)| \gt |g(x_1) - g(x_2)|.$$

By factorizing we have $$|e^{x_1} - e^{x_2}| \gt |(x_1 - x_2)(x_1 + x_2 + a)|.$$ But I am not sure how to proceed. In particular, I don't know how to deal with the absolute values and the two variables. Assume that $x_1 \gt x_2$ (without loss of generality) and we have $$\frac {e^{x_1} - e^{x_2}}{x_1 - x_2} \gt |x_1 + x_2 + a|,$$ but that doesn't seem to yield any solution.
This question is about derivatives, if that helps.
 A: Fix $x \in [0, 2]$, under the given condition, it follows that for $t > x$, 
$$\left|\frac{e^t - e^x}{t - x}\right| > \left|\frac{g(t) - g(x)}{t - x}\right|. \tag{1}$$
Let $t \downarrow x$, we then have (treat boundary points similarly)
$$e^x \geq \left|g'(x)\right| = \left|2x + a\right| \quad \text{ for all } x \in [0, 2] \tag{2}$$
Geometrically, $(2)$ implies that the graph of $y = e^x$ always sits at the top of the graph of $y = |2x + a|$. 
If $a \geq 0$, since $x = \log 2 \in [0, 2]$ and $e^{\log 2} = 2$, the intercept of the line $y - 2 = 2(x - \log 2)$ is $2 - 2\log 2 > 0$, thus $a \in [0, 2 - 2\log 2]$.
If $a < 0$, on $(0, +\infty)$, $y = |2x + a|$ has a ''V'' shape, with the intersection with the $x$-axis at $-a/2$, and the intersection with $y$-axis at $-a$. If $-a/2 \geq 2$, then $-a \leq 1$ guarantees $(2)$, but these two conditions are inconsistent. Thus we consider the case $-a/2 < 2$, for which case $-a \leq 1$ and $4 + a \leq e^2$ ensure $(2)$. Solve these inequalities gives $a \geq -1$. 
In summary, we conclude that if $a \in [-1, 2 - 2\log 2]$ then we can guarantee $(2)$. We need also to check for $a \in [-1, 2 - 2\log 2]$, we indeed have $|g(x_2) - g(x_1)| < |f(x_2) - f(x_1)|$. In fact, in view of $(2)$, for $x_2 > x_1$, if $a \in [0, 2 - 2\log 2]$, then
\begin{align}
& |f(x_2) - f(x_1)| = e^{x_2} - e^{x_1} \\
= & \int_{x_1}^{x_2} e^t dt \\
\geq & \int_{x_1}^{x_2} |2t + a| dt \quad \text{because $(2)$ holds, given $a \in [0, 2 - 2\log 2]$}\\
= & \int_{x_1}^{x_2} (2t + a) dt \\
= & x_2^2 - x_1^2 + a(x_2 - x_1) = |g(x_2) - g(x_1)|.
\end{align}
If $a \in [-1, 0)$, for which case we should discuss where $x_1$ and $x_2$ located, if $\max\{x_1, x_2\} < -a/2$, then
\begin{align}
& \int_{x_1}^{x_2} |2t + a| dt \\
= & -\int_{x_1}^{x_2} (a + 2t) dt \\
= & -a(x_2 - x_1) - (x_2^2 - x_1^2).
\end{align}
Note that since $x_1 + x_2 < -a$, the last expression is positive and equals to $|g(x_2) - g(x_1)|$. The case of $\min\{x_1, x_2\} \geq -a/2$ can be treated in a similar manner. For the case $x_1 < -a/2 < x_2$:
\begin{align}
& \int_{x_1}^{x_2} |2t + a| dt \\
= & \int_{x_1}^{-a/2} |2t + a| dt + \int_{-a/2}^{x_2}|2t + a| dt \\
= & -\int_{x_1}^{-a/2} (a + 2t) dt + \int_{-a/2}^{x_2}(2t + a) dt \\
= & x_1^2 + x_2^2 + a(x_2 + x_1) + a^2/2
\end{align}
Careful calculation shows that 
\begin{align}
& x_1^2 + x_2^2 + a(x_2 + x_1) + a^2/2 - |g(x_2) - g(x_1)| \\
= & \frac{1}{2}(2x_1 + a)^2I(x_1 + x_2 > -a) + \frac{1}{2}(2x_2 + a)^2I(x_1 + x_2 \leq -a) \geq 0.
\end{align}
We therefore showed that if $a \in [-1, 2 - 2\log 2]$, then $|f(x_2) - f(x_1)| > |g(x_2) - g(x_1)|$. If you need strict inequality, we can set $a \in (-1, 2 - 2\log 2)$, for safety.
A: HINT: You have almost reached your solution.
$$\frac {e^{x_1} - e^{x_2}}{x_1 - x_2} \gt |x_1 + x_2 + a|$$
$$\Rightarrow  |x_1 + x_2 + a| \lt \frac {e^{x_1} - e^{x_2}}{x_1 - x_2}$$
$$\Rightarrow  -\frac {e^{x_1} - e^{x_2}}{x_1 - x_2} \lt x_1 + x_2 + a \lt \frac {e^{x_1} - e^{x_2}}{x_1 - x_2}$$
$$\Rightarrow  -\frac {e^{x_1} - e^{x_2}}{x_1 - x_2}-(x_1 + x_2)  \lt a \lt \frac {e^{x_1} - e^{x_2}}{x_1 - x_2}-(x_1 + x_2)$$
So $a$ lies in the interval $\{-\frac {e^{x_1} - e^{x_2}}{x_1 - x_2}-(x_1 + x_2),\frac {e^{x_1} - e^{x_2}}{x_1 - x_2}-(x_1 + x_2)\}$
Now you just need to apply one more condition to this solution to get the correct answer: $$x_1,x_2 \in [0,2]$$
And this can be done without derivatives.
Can you complete the answer now?
P.S. However if you seek to do it the derivatives way, try to use Lagrange's Mean Value Theorem. Then we will get the same solution.
