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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

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  • $\begingroup$ Graph the function. Observe. $\endgroup$ Nov 2, 2016 at 19:33
  • $\begingroup$ Verify that $a \le b \iff \sqrt a \le \sqrt b$. $\endgroup$
    – dxiv
    Nov 2, 2016 at 19:34
  • $\begingroup$ If you know what the graph looks like, then you can determine on which parts of the domain the function is increasing by taking your pencil and outlining/tracing the graph of the function from left to right. When your pencil is moving upward, the function is increasing. When your pencil is moving downward, the function is decreasing. If your pencil ever moves horizontally, then the function is neither increasing nor decreasing there. $\endgroup$
    – layman
    Nov 2, 2016 at 19:35
  • $\begingroup$ I added an edit in my answer to show you the specific $\sqrt{x}$ example! $\endgroup$
    – Enrico M.
    Nov 2, 2016 at 20:10

2 Answers 2

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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.

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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.

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  • $\begingroup$ So would you usually differentiate it then see if the gradients always positive or graph it to see when it's increasing? $\endgroup$
    – Thomas
    Nov 2, 2016 at 19:46

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