Sometimes it is easy to see exactly where a function has its highest or lowest values just by looking at its graph. Most often, though, the answer requires analytical work.
I am starting this new Blog with an example on how to find analytically on what interval(s) a function is increasing or decreasing. For this I will use the First Derivative Test.
The function, and its graph are: \[f(x)=\frac{1}{2}x-sin(x)\tag{1}\]
For this, we need to:
- Take the derivative of the function: \[f'(x)=\frac{1}{2}-cos(x)\tag{2}\]
- Set the derivative function equal to zero to find our where the critical numbers are:
\begin{alignat*}{1}
\frac{1}{2}-\cos(x)&=0\\
\cos(x)&=\frac{1}{2}\\
x&=\frac{\pi}{3},\frac{5\pi}{3}
\end{alignat*} - The Critical Numbers are \(\frac{\pi}{3}\) and \(\frac{5\pi}{3}\)
- The last step is to test intervals: \((0,\frac{\pi}{3})\), \((\frac{\pi}{3},\frac{5\pi}{3})\) and \((\frac{5\pi}{3},2\pi)\).
The following table demonstrates this technique:
| Interval | \(0<x<\frac{\pi}{3}\) | \(\frac{\pi}{3}<x<\frac{5\pi}{3}\) | \(\frac{5\pi}{3}<x<2\pi\) |
| Test Value | \(x=\frac{\pi}{4}\) | \(x=\pi\) | \(x=\frac{7\pi}{4}\) |
| Sign of \(f'(x)\) | \(f'(\frac{\pi}{4})<0\) | \(f'(\pi)>0\) | \(f'(\frac{7\pi}{4})<0\) |
| Conclusion | Decreasing | Increasing | Decreasing |
As the graph shows, the \(f(x)\) function is decreasing from \(0\) to \(\frac{\pi}{3}\); increasing to \(\frac{5\pi}{3}\), and finally decreasing to \(2\pi\).
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .
I hope to do a calculus post every week.