Proving a increasing function with algebra

Question

I'm attempting to prove a quadratic function is increasing without any calculus, just using algebra facts. My question:

Consider the function $g(x) = (x + \dfrac{1}{2})^2 + \dfrac{7}{4}$

Prove that For every $x_1$ and $x_2$ in the interval [2,4], IF $x_2$ > $x_1$, THEN $g(x_2)$ > $g(x_1)$

This is what I've done so far:

$$(x_2 + \dfrac{1}{2})^2 > (x_1 + \dfrac{1}{2})^2$$ $$(x_2 + \dfrac{1}{2})^2 + \dfrac{7}{4} > (x_1 + \dfrac{1}{2})^2 + \dfrac{7}{4}$$ $$∴ g(x_2) > g(x_1)$$

Does that actually prove anything? Because by basic algebra that if $x_2$ is greater than $x_1$, the square of that will be bigger no matter what in the interval [2,4].

Also need to show what goes wrong with my proof if I instead use the interval [-2,4]

Answer 1

Your proof is not quite right; what I'd do is notice that we can clearly move from $x_2>x_1$ to $$x_2+\frac{1}2>x_1+\frac{1}2,$$ but, unless both sides are positive, squaring might not preserve the order - for instance, if we had $1>-2$ and squared both sides, we'd get $1>4$, which is wrong. So, before you write that $$(x_2+\frac{1}2)^2>(x_1+\frac{1}2)^2,$$ you need to check that both sides are positive. They are both positive if both $x$ are in $[2,4]$, meaning that the function is increasing in that interval, but if that range is extended to $[-2,4]$, it is possible to make one or both sides negative. In particular, if you choose any two $x_1,x_2<\frac{-1}2$, both sides are negative, and for $x_2>x_1$, it will be that $$(x_2+\frac{1}2)^2<(x_1+\frac{1}2)^2$$ since when you square an inequality with negatives on both sides, you have to flip the direction of the inequality. So you could choose any counterexample from that range.

Answer 2

First, plot the function:

You can guess from the plot and the form of the function that $g$ is non-decreasing on $[-{1 \over 2}, \infty)$.

So, suppose $-{1 \over 2} \le x_1 \le x_2$. Then $0 \le x_1 + {1 \over 2} \le x_2 + {1 \over 2}$, and since the function $x \mapsto x^2$ is increasing on $[0,\infty)$ (this is the key fact here), we have $(x_1 + {1 \over 2})^2 \le (x_2 + {1 \over 2})^2$. It follows from this that $g(x_1) \le g(x_2)$.

The $\ge {1 \over 2}$ is the important restriction in the proof. Without this constraint, we cannot use the fact that $x \mapsto x^2$ is increasing.

In the interval you asked about, $[-2,4]$, we can see that we can find $x_1,x_2 \in [-2,4]$ so that $x_1 < x_2$ and $(x_1 + {1 \over 2})^2 > (x_2 + {1 \over 2})^2$ (and so $g(x_1) > g(x_2)$). The reasoning above shows that we must have $x_1 < - {1 \over 2}$.

A quick look at the graph shows that $g(-2) > g(0)$ and a quick computation verifies this.

Answer 3

Yes, you've shown that the function on the interval $[2, 4]$ is increasing.

To see that $g(x)$ is not strictly increasing on all of $[-2, 4]$ take, e.g., $x_2 = 1 \gt x_1 = -2$.

We have $x_2 \gt x_1$, but $g(x_2) \lt g(x_1)$. Can you see why $g(x)$ is increasing for all $x \in \left(-\frac 12, +\infty\right)$, neither increasing nor decreasing at $x = -\frac 12$, and decreasing for all $x \in (-\infty, -1/2)$?

Answer 4

All your functions (+, square) are monotonically increasing on the interval [2, 4], so their composition is monotonic too.

This speculation fails on the interval [-2, 4], because summation with constant is still monotonically increasing, but that can't be said about the square function.