I'm getting better at using trig functions, but this problem has me up against a wall.

As I understand it, to find if a function is increasing or decreasing over a given interval, select a number inside that interval and if the result is greater or less than zero, the interval is increasing and decreasing, respectively.

Given $f(x)=5x+10 \sin x$ on the interval $(0, 2\pi)$, find the open intervals where the function is increasing or decreasing.

The given intervals are $(0, \frac{2\pi}{3}), (\frac{2\pi}{3}, \frac{4\pi}{3}), (\frac{4\pi}{3}, 2\pi).$

On the first interval, I selected $\frac{\pi}{2}$ as the test for x. Thus $$f(x)= 5x+10 \sin x \Rightarrow 5(\frac{\pi}{2}) + 10 \sin(\frac{\pi}{2})\Rightarrow \frac{5\pi}{2}+10$$

$\frac{5\pi}{2}+10 > 0$, so the function should be increasing at that point

The second interval test variable was $\pi$, and thus $5(\pi)+10 \sin(\pi)= 5\pi > 0$ So the interval of $(\frac{2\pi}{3}, \frac{4\pi}{3})$ should be increasing.

The third interval test variable was $\frac{3\pi}{2}$, and $5(\frac{3\pi}{2})+10 \sin(\frac{3\pi}{2})= \frac{15\pi}{2}-10 > 0$, so that interval is increasing as well.

However, the WebAssign site didn't accept those answers. What is missing with my logic and work?

  • $\begingroup$ Concerning the title of your question: the concept of "increasing" doesn't really make sense at a single point, since it involves comparing $f(a)$ to $f(b)$, where $a$ and $b$ are different points. (The function $f$ is increasing on an interval if $f(a)<f(b)$ for every pair of points $a<b$ in that interval.) $\endgroup$ – Hans Lundmark Mar 26 '11 at 19:22

You are nearly on the right track but are missing a key ingredient.

In calculus we learn that if the DERIVATIVE of a function is positive on an interval, then the function is increasing on that interval. If the DERIVATIVE of a function is negative on an interval, then the function is decreasing on that interval. So, you should first take the derivative to get $f'(x) = 5 + 10\cos x$ and then plug in values to the derivative (not the original function). That is, $f'(\pi/2) = 5 > 0$ so $f$ is increasing on $(0, 2\pi/3)$. $f'(\pi) = -5 < 0$, so $f$ is decreasing on the interval $(2\pi/3,4\pi/3)$. $f'(3\pi/2) = 5 > 0$, so $f$ is increasing on the third interval.

Now, there is also another missing step, which is to find the correct intervals to use. It seems as if you are given these intervals in the problem. If they were not given, or if you want to ensure you were given the correct intervals, you find where the derivative is 0 or undefined.


On a given interval, a function can be increasing, decreasing, or neither; "neither" can, however, mean that there are some places in the interval where the function is increasing and some places where the function is decreasing. To test whether a function is increasing (or decreasing) on an interval, you have to check whether the function is increasing (or decreasing) everywhere in that interval.

For your specific problem, I'd start by considering where the derivative of your function is positive and where it is negative (or, perhaps more simply, where is the derivative zero?).

  • $\begingroup$ so if there is a derivative value of zero falling within an interval, it will mean the interval is neither? $\endgroup$ – Jason Mar 26 '11 at 18:54
  • $\begingroup$ @Jason: In general, not necessarily (think about $f(x)=x^5$, which is increasing everywhere, but has $f'(0)=0$), but that's definitely a good place to start. (For your particular function, do you think that there is any "unusual" behavior like $x^5$ has near 0?) $\endgroup$ – Isaac Mar 26 '11 at 18:58
  • $\begingroup$ @Jason: No. For example, the function $y=x^3$ has derivative $y'=3x^2$, which is $0$ at $0$, but the function is increasing on $[-1,1]]$ anyway. $\endgroup$ – Arturo Magidin Mar 26 '11 at 19:16

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