Question

(a) Determine the intervals on which the function \(x = \frac{\pi }{4},\;x = \frac{{5\pi }}{4}\) on \((0,\;2\pi )\) is increasing or decreasing.

(b) Determine the local maximum and minimum values of \(f(x) = \sin x + \cos x\).

(c) Determine the intervals of concavity and the inflection points of\(f(x) = \sin x + \cos x\).

Short answer

(a) The function \(f(x)\) is increasing on \(\left( {0,\;\frac{\pi }{4}} \right) \cup \left( {\frac{{5\pi }}{4},\;2\pi } \right)\) and decreasing on \(\left( {\frac{\pi }{4},\;\frac{{5\pi }}{4}} \right)\).

(b) The local maximum is \(f\left( {\frac{\pi }{4}} \right) = \sqrt 2 \) and local minimum is \(f\left( {\frac{{5\pi }}{4}} \right) = - \sqrt 2 \).

(c) The function \(f(x)\) is concave upward on \(\left( {\frac{{3\pi }}{4},\frac{{7\pi }}{4}} \right)\), concave downward on \(\left( {0,\frac{{3\pi }}{4}} \right) \cup \left( {\frac{{7\pi }}{4},2\pi } \right)\) and the inflection points are \(\left( {\frac{{3\pi }}{4},0} \right)\) and \(\left( {\frac{{7\pi }}{4},0} \right)\).

Step-by-step solution

Given data

The given function is, \(f(x) = \sin x + \cos x\).

Concept of Increasing/decreasing test and concavity

Increasing/decreasing test:

(a) If\({f^\prime }(x) > 0\)on an interval, then\(f\)is increasing on that interval.

(b) If\({f^\prime }(x) < 0\)on an interval, then\(f\)is decreasing on that interval.

Definition of concavity:

"(a) If\({f^\prime }(x) > 0\)for all\(x\)in\(I\), then the graph of\(f\)is concave upward on\(I\).

(b) If \({f^\prime }(x) < 0\)for all\(x\)in\(I\), then the graph of\(f\)is concave downward on\(I\).”

Obtain the critical points 

(a)

Obtain the derivative of \(f(x)\) as follows:

\(\begin{aligned}{c}{f^\prime }(x) &= \frac{d}{{dx}}(\sin x + \cos x)\\ &= \cos x - \sin x\end{aligned}\)

Set \({f^\prime }(x) = 0\) and obtain the critical points as follows:

\(\begin{aligned}{c}\cos x - \sin x &= 0\\\cos x &= \sin x\\\frac{{\sin x}}{{\cos x}} &= 1\\\tan x &= 1\end{aligned}\)

The trigonometry function, \(\tan x = 1\) when, \(x = \frac{\pi }{4},\;x = \frac{{5\pi }}{4}\).

Hence, the critical points occurs at \(x = \frac{\pi }{4},\;x = \frac{{5\pi }}{4}\) and in the interval notation\(\left( {0,\;\frac{\pi }{4}} \right) \cup \left( {\frac{\pi }{4},\;\frac{{5\pi }}{4}} \right) \cup \left( {\frac{{5\pi }}{2},\;2\pi } \right)\).

Determine the nature of \(f(x)\)

Determine the nature of \(f(x)\)in table 1 as follows:

Table 1

From the above table observe that, \(f(x)\) is increasing on \(\left( {0,\;\frac{\pi }{4}} \right) \cup \left( {\frac{{5\pi }}{2},\;2\pi } \right)\) and \(f(x)\) is decreasing on\(\left( {\frac{\pi }{4},\;\frac{{5\pi }}{4}} \right)\).

Determine the local maximum and local minimum

(b)

From part (a), the function \(f(x)\) is decreasing on \(\left( {\frac{\pi }{4},\frac{{5\pi }}{4}} \right)\).

Thus, the local maximum occurs at \(x = \frac{\pi }{4}\) and the local minimum occurs at \(x = \frac{{5\pi }}{4}\)

Substitute \(x = \frac{\pi }{4}\) in \(f(x)\) and obtain the local maximum as follows:

\(\begin{aligned}{c}f\left( {\frac{\pi }{4}} \right) &= \sin \left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{\pi }{4}} \right)\\ &= \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}\\ &= \frac{2}{{\sqrt 2 }}\\ &= \sqrt 2 \end{aligned}\)

Substitute \(x = \frac{{5\pi }}{4}\) in \(f(x)\) and obtain the local minimum as follows:

\(\begin{aligned}{c}f\left( {\frac{{5\pi }}{4}} \right) &= \sin \left( {\frac{{5\pi }}{4}} \right) + \cos \left( {\frac{{5\pi }}{4}} \right)\\ &= \frac{{ - 1}}{{\sqrt 2 }} + \frac{{ - 1}}{{\sqrt 2 }}\\ &= \frac{{ - 2}}{{\sqrt 2 }}\\ &= - \sqrt 2 \end{aligned}\)

Thus, the local maximum is \(f\left( {\frac{\pi }{4}} \right) = \sqrt 2 \) and the local minimum is \(f\left( {\frac{{5\pi }}{4}} \right) = - \sqrt 2 \).

Determine the nature of \(f(x)\)

Determine the nature of \(f(x)\) based on the intervals of \(x\) in table 2 as follows:

Table 2

Hence, \(f(x)\) is concave upward on \(\left( {\frac{{3\pi }}{4},\;\frac{{7\pi }}{4}} \right)\) and concave downward on \(\left( {0,\;\frac{{3\pi }}{4}} \right) \cup \left( {\frac{{7\pi }}{4},\;2\pi } \right)\).

Hence, the curve changes from concave downward to concave upward at \(x = \frac{{3\pi }}{4}\) and \(x = \frac{{7\pi }}{4}\).

Substitute \(x = \frac{{3\pi }}{4}\) in \(f(x)\) and find the inflection point as follows:

\(\begin{aligned}{c}f\left( {\frac{{3\pi }}{4}} \right) &= \sin \left( {\frac{{3\pi }}{4}} \right) + \cos \left( {\frac{{3\pi }}{4}} \right)\\ &= \frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}\\ &= \frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}\\ &= 0\end{aligned}\)

Substitute \(x = \frac{{7\pi }}{4}\) in \(f(x)\) and find the inflection point as follows:

\(\begin{aligned}{c}f\left( {\frac{{7\pi }}{4}} \right) &= \sin \left( {\frac{{7\pi }}{4}} \right) + \cos \left( {\frac{{7\pi }}{4}} \right)\\ &= \frac{{ - \sqrt 2 }}{2} + \frac{{\sqrt 2 }}{2}\\ &= \frac{{ - 1}}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}\\ &= 0\end{aligned}\)

Thus, the inflection points are \(\left( {\frac{{3\pi }}{4},\;0} \right)\) and \(\left( {\frac{{7\pi }}{4},\;0} \right)\).