How to determine if a function is increasing. 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
 A: 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.
A: 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.
