# What cases to consider when proving that $x_1 < x_2 \implies f(x_1) \leq f(x_2)$, i.e. function is increasing

So I have the following function $f$: $$f(x) = \begin{cases} 12x - 4100 && \text{if x < - 2}\\ (x - 14)^3 - 21 && \text{if x \geq - 2} \end{cases}$$ Okay, so I want to prove that this function is increasing by using monotonicity and no derivatives. Do I have to prove the following 3 cases? I'm not sure about case 3, how do I know what to put in my case 3?

Case 1 : $x_1,x_2 < -2$

Case 2 : $x_1,x_2 \geq -2$

Case 3 : $x_1 \leq -2 \leq x_2$

• In case 3, take $x_1 < -2 \leq x_2$. – N. F. Taussig Nov 12 '17 at 22:16
• @N.F.Taussig , With your correction of Case 3, have I done it correct by doing the following? 12x1 - 4100 < 12(-2) - 4100 < (-2-14)^3 - 21 <= (x2 -14)^3 - 21 – Lukas Nov 12 '17 at 22:31
• Your inequalities are correct. However, you should state, or better yet prove, that $12x - 4100$ is monotonically increasing to justify the first inequality and that $(x - 14)^3 - 21$ is monotonically increasing to justify the last inequality. – N. F. Taussig Nov 12 '17 at 22:35
• @N.F.Taussig , hmm isn't that what I'm supposed to do in case 1 and case 2? So for case 1: 12x1 - 4100 < 12x2 - 4100 and for case 2: (x1-14)^3 - 21 < (x2-14)^3 - 21 – Lukas Nov 12 '17 at 22:46

For case 3, we take $x_1 < -2 \leq x_2$.
Since $x_1 < -2$, $-2 - x_1 > 0$. Hence, $$12(-2) - 4100 - [12(x_1) - 4100] = 12(-2 - x_1) > 0 \implies 12x_1 - 4100 < 12(-2) - 4100$$ as you claimed in the comments.
Direct calculation shows that $12(-2) - 4100 = -4124 < -4117 = (-2 - 14)^3 - 21$.
Since $x_2 \geq -2$, $x_2 - (-2) \geq 0$.
\begin{align*} f(x_2) - f(-2) & = (x_2 - 14)^3 - 21 - [(-2 - 14)^3 - 21]\\ & = x_2^3 - 42x_2^2 + 588x_2 - 2744 - 21 - [(-2)^3 - 42(-2)^2 + 588(-2) - 2744 - 21]\\ & = [x_2^3 - (-2)^3] - 42[x_2^2 - (-2)^2] + 588[x_2 - (-2)]\\ & = [x_2 - (-2)][x_2^2 + (-2)x_2 + (-2)^2] - 42[x_2 - (-2)](x_2 + (-2)] + 588[x_2 - (-2)]\\ & = [x_2 - (-2)][x_2^2 + (-2)x_2 + (-2)^2 - 42x_2 - 42(-2) + 588x_2 - 588(-2)]\\ & = [x_2 - (-2)]\{x_2^2 + 544[x_2 - (-2)]\}\\ & \geq 0 \end{align*} since $x_2 - (-2) \geq 0$ and $x_2^2 + 544[x_2 - (-2)] \geq 0$.
Therefore, $$f(x_1) = 12x_1 - 4100 < 12(-2) - 4100 < (-2 - 14)^3 - 21 \leq (x_2 - 14)^3 - 21 = f(x_2)$$ which proves the result holds for case 3.