Find the intervals in which f(x)=sin x+ cos x, 0 leq x leq 2pi is strictly increasing/decreasing.

princetonaqo3

princetonaqo3

Answered question

2022-10-12

Intervals in which f(x) is Strictly Increasing/Decreasing
Find the intervals in which f ( x ) = sin x + cos x , 0 x 2 π is strictly increasing/decreasing.
First I find the derivative f ( x ) = cos x sin x, then put f ( x ) = 0, getting tan x = 1. The principal solutions of tan x = 1 are x = π / 4 and x = 5 π / 4, which gives the intervals [ 0 , π / 4 ), ( π / 4 , 5 π / 4 ), and ( 5 π / 4 , 2 π ).
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 ( π / 4 , 5 π / 4 ).

Answer & Explanation

Taxinov

Taxinov

Beginner2022-10-13Added 18 answers

Step 1
The function f′(x) is continuous in [ 0 , 2 π ]. We have f ( x ) = 0 at x = π 4 and x = 5 π 4 .
Then, it can be concluded that f′(x) is either positive for all x in ( π 4 , 5 π 4 ) or negative for all x in ( π 4 , 5 π 4 ). 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 = π 2 , f ( x ) = 1.
Thus, f′(x) is negative in ( π 4 , 5 π 4 ) and it can be concluded that f(x) is strictly decreasing in this interval.

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