How to determine if a function is increasing.

Question

How would I determine whether a function is increasing, decreasing or neither without using calculus? Like whether x^0.5 is increasing in interval [0, infinity) Just curious Thanks

Answer 1

There are many non-calculus techniques that can be applied to show a function is increasing. One approach is to show for $k > 0$ that $f(x + k)$ is larger than $f(x)$.

Here’s a simple example.

Suppose that $f(x) = mx + b$ Now take $k > 0$ and compare $f(x)$ to $f(x+k)$.

$$f(x+k) - f(x) = mk$$

Since $k > 0$ we know that $mk$ is positive iff $m$ is positive, hence $f(x)$ is increasing if $m$ is positive.

Another example

Take $f(x) = x^3$

$$f(x+k) = (x+k)^3 = x^3 + k^3 + 3k^2x + 3x^2k$$

$$f(x+k) - f(x) = 3kx^2 + 3k^2x + k^3$$

For $x > 0$ all terms are positive, hence the function is increasing.

At $x = 0$ the difference is $k^3$ which is also positive by definition.

For $x < 0$ we can take $k$ arbitrarily small (but positive) to show that $f(x)$ is still increasing.

Let $k < |x|$:

$$f(x+k) - f(x) = 3kx^2 + 2k^2x + k^3$$

$$f(x+k) - f(x) = k(3x^2 +3kx) + k^3 > k(3x^2 + 3kx) > k(3x^2 + 3|x|x)$$

Since we said $x$ is negative we have

$$f(x+k) - f(x) = k(3x^2 - 3x^2) = 0$$

The difference is positive, hence the function is increasing.

Now you can try this method with $\sqrt{x}$

$\sqrt{x}$

$$f(x+k) = \sqrt{x+k}$$

$$f(x+k) - f(x) = \sqrt{x+k} - \sqrt{x}$$

To solve it, lets' multiply and divide by

$$\frac{\sqrt{x+k} + \sqrt{x}}{\sqrt{x+k} + \sqrt{x}}$$

So we have

$$(\sqrt{x+k} - \sqrt{x})\cdot \frac{\sqrt{x+k} + \sqrt{x}}{\sqrt{x+k} + \sqrt{x}} = \frac{k}{\sqrt{x+k} + \sqrt{x}}$$

Since $k> 0$ and since the square root admits only positive real numbers (in $\mathbb{R}$), and since the denominator is a sum of positive terms, you have that

$$ \frac{k}{\sqrt{x+k} + \sqrt{x}}$$

Is always positive, hence the function $\sqrt{x}$ is increasing.

Answer 2

To prove that $f$ is increasing, prove that whenever $x \le y$, we have $f(x)\le f(y)$ (this is the definition by the way)

Ex: for $f:x \mapsto \sqrt x$

Let $x \le y$. Then, $0\le (\sqrt y)^2 -(\sqrt x)^2 = (\sqrt y - \sqrt x)(\sqrt y + \sqrt x)$. So $\sqrt y- \sqrt x\ge 0$, then $\sqrt y \ge \sqrt x$ i.e. $f(x)\le f(y)$.

However, this was a very simple example. For almost any function you'll meet, it will be tedious (if not impossible) to study its sense starting from the definition.