# strictly increasing function

• Mar 28th 2009, 11:59 PM
profmq
strictly increasing function

we are currently doing functions in 1st year university, and often in order to justify why a function has an inverse, we must restrict the domain to the largest interval for which the function is strictly increasing (or decreasing).

now, my lecturer defined 'strictly increasing' as 'if x1 < x2, f(x1) < f(x2)'.

now in order to find where a function is strictly increasing, i find the derivative and find for what domain it is greater than 0 for.

however, my lecturer gave us an example, saying for y=sinx, it is strictly increasing over [-pi/2, pi/2] ---> this is a CLOSED INTERVAL

So as i do problems now, i am having trouble identifying whether the stationary points should be included in the largest interval for which f(x) is strictly increasing (as my lecturer said for sinx), or not

ie. if a question asks to find over what interval a function is strictly increasing, do we include the stationary points (ie make it a closed interval, beacuse technically x1 < x2, f(x1) < f(x2) still holds, however, if finding it by finding when f'(x)>0, you dont get those points???

could someone clarify this further by telling me the largest interval in which this function is strictly increasing : f(x) = x/(1+x^2)

is it (-1,1) or [-1,1]

thankyou.
• Mar 29th 2009, 01:20 AM
running-gag
Hi

The function $f(x) = \frac{x}{1+x^2}$ is strictly increasing over [-1,1]

Of course the derivative is equal to 0 for x=-1 and x=1 but due to the fact that these are the boundaries of the interval and that f is continuous, it makes no sense to say that f is strictly increasing over (-1,1)

If you come back to the definition
Let $x_0 < 1$
$f(x_0)-f(1) = \frac{x_0}{1+x_0^2} - \frac12 = \frac{-x_0^2+2x_0-1}{2(1+x_0^2)} = -\frac{(x_0-1)^2}{2(1+x_0^2)}$

Therefore $f(x_0)-f(1) < 0$ or $f(x_0) < f(1)$
• Mar 29th 2009, 01:29 AM
profmq
okay thanks for that

so if i am asked to find where a fuinction is strictly increasing over, i use the closed interval, ie when f'(x)>=0??

because some websites (like wikipedia) said 'strictly increasing' meant f'(x)>0 (no equality), thats what sort of confused me....
• Mar 29th 2009, 01:40 AM
running-gag
If f is defined over a closed interval and f'(x) > 0 on the open interval then f is strictly increasing over the closed interval (because f is continuous) even if f = 0 on the boundaries of the interval

If f'(x) = 0 on a point inside the domain of f it depends by the sign of f' before and after this point

Let 's take an example
$f(x) = x^3$

$f'(x) = 3x^2$

f'(x) > 0 over $]-\infty,0[$ and > 0 over $]0,+\infty[$ but f'(0)=0

However f is strictly increasing over $]-\infty,+\infty[$