princetonaqo3

2022-10-12

Intervals in which f(x) is Strictly Increasing/Decreasing
Find the intervals in which $f\left(x\right)=\mathrm{sin}x+\mathrm{cos}x,0\le x\le 2\pi$ is strictly increasing/decreasing.
First I find the derivative ${f}^{\prime }\left(x\right)=\mathrm{cos}x-\mathrm{sin}x$, then put ${f}^{\prime }\left(x\right)=0$, getting $\mathrm{tan}x=1$. The principal solutions of $\mathrm{tan}x=1$ are $x=\pi /4$ and $x=5\pi /4$, which gives the intervals $\left[0,\pi /4\right)$, $\left(\pi /4,5\pi /4\right)$, and $\left(5\pi /4,2\pi \right)$.
After this I am stuck. The book just solves the question by making a table showing the interval, sign of the derivative as positive or negative, and strictly increasing for positive and vice versa.
I just want to know how do they get it positive or negative.It would really help if someone did one for the interval $\left(\pi /4,5\pi /4\right)$.

Taxinov

Step 1
The function f′(x) is continuous in $\left[0,2\pi \right]$. We have ${f}^{\prime }\left(x\right)=0$ at $x=\frac{\pi }{4}$ and $x=\frac{5\pi }{4}$.
Then, it can be concluded that f′(x) is either positive for all x in $\left(\frac{\pi }{4},\frac{5\pi }{4}\right)$ or negative for all x in $\left(\frac{\pi }{4},\frac{5\pi }{4}\right)$. This follows from the continuity of f′(x) and the intermediate value theorem.
Step 2
Now, all that is left to find is whether it takes positive or negative values. This can be done easily by checking the value of f′(x) at any convenient x in the interval.
For example, at $x=\frac{\pi }{2}$, ${f}^{\prime }\left(x\right)=-1$.
Thus, f′(x) is negative in $\left(\frac{\pi }{4},\frac{5\pi }{4}\right)$ and it can be concluded that f(x) is strictly decreasing in this interval.

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