When the graph of a function rises from left to right, we say the function *increases*. Similarly, when the graph falls from left to right, we say the function *decreases*. The following definitions make these terms more precise:

Suppose a function $f$ is defined on some interval $I$. Then,

$f$ is

**increasing on $I$**if for every pair of values $x_1 \lt x_2$ in $I$, $f(x_1) \lt f(x_2)$.$f$ is

**decreasing on $I$**if for every pair of values $x_1 \lt x_2$ in $I$, $f(x_1) \gt f(x_2)$.$f$ is

**monotonic on $I$**if $f$ is either increasing on $I$ or decreasing on $I$.

Now suppose our function is continuous on $[a,b]$ and differentiable on $(a,b)$. Further, let both $x_1$ and $x_2$ be any two $x$-values in $[a,b]$ with $x_1 \lt x_2$. Note the conditions of the mean value theorem relative to the interval $[x_1,x_2]$ are satisfied, so the mean value theorem applies and concludes there exists some $c$ in $(x_1,x_2)$ where

$$f'(c) = \frac{f(x_2)-f(x_1)}{x_2-x_1}$$But then, $f'(c)(x_2-x_1) = f(x_2)-f(x_1)$, and upon solving for $f(x_2)$ we have

$$f(x_2) = f(x_1) + f'(c)(x_2-x_1)$$Of course, as $x_1 \lt x_2$, it must be the case that $(x_2-x_1)$ is positive.

So if $f'(c) \gt 0$, we know $f(x_1) \lt f(x_2)$, whereas if $f'(c) \lt 0$, it must instead be the case that $f(x_1) \gt f(x_2)$.

Applying the definitions of increasing and decreasing above, we can then establish the following important result:

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then:

- If $f'(x) \gt 0$ for all $x$ in $(a,b)$, then $f$ is increasing on $[a,b]$;
- If $f'(x) \lt 0$ for all $x$ in $(a,b)$, then $f$ is decreasing on $[a,b]$;

An immediate application of the above helps us prove the following important test for finding certain local minimums and maximums of a function:

__The First-Derivative Test__

Suppose $f$ is a function continuous on $(a,b)$, where $c$ is some point in this interval. Further presume that $f$ is differentiable at all points of $(a,b)$, except possibly at $c$. Then,

If $f'(x)$ is positive for all values $x$ in $(a,c)$ and negative for all values $x$ in $(c,b)$,

then $f$ has a local maximum at $c$.If $f'(x)$ is negative for all values $x$ in $(a,c)$ and positive for all values $x$ in $(c,b)$,

then $f$ has a local minimum at $c$.

To see this, consider the following argument to establish part (*i*):

Let $a_0$ be the midpoint between $a$ and $c$. Then $f$ is continuous on $[a_0,c]$ and differentiable on $(a_0,c)$. If $f'(x)$ is positive for all $x$ in $(a,b)$, then $f'(x) \gt 0$ for all $x$ in $(a_0,c)$. Thus, by the aforementioned result, $f$ is increasing on $[a_0,c]$. So for any $x$ in $(a_0,c)$, we have $f(x) \lt f(c)$.

Similarly, let $b_0$ be the midpoint between $c$ and $b$. Then $f$ is continuous on $[c,b_0]$ and differentiable on $(c,b_0)$. If $f'(x)$ is negative for all $x$ in $(a,b)$, then $f'(x) \lt 0$ for all $x$ in $(c,b_0)$. Thus, by the aforementioned result, $f$ is decreasing on $[c,b_0]$. So for any $x$ in $(c,b_0)$, we have $f(x) \lt f(c)$.

Putting these two conclusions together tells us that for any $x$ in the open interval $(a_0,b_0)$, we have $f(x) \le f(c)$. Thus, $f$ has a local maximum at $c$.

Part (*ii*) of the First Derivative Test is established in a similar fashion.