Question

Find intervals of concavity and points of inflection for the following functions:

f(x) = sin x + cos x, 0 < x < 2π

Answer 1

f'(x) = cos x – sin x

f”(x) = – sin x – cos x

f'(x) = 0

⇒ sin x + cos x = 0

Critical points x = ${\displaystyle \frac{3\pi }{4},\frac{7\pi }{4}}$

The intervals are ${\displaystyle (0,\frac{\pi }{4}),(\frac{3\pi }{4},\frac{7\pi }{4})}$ and ${\displaystyle (\frac{7\pi }{4},2\pi )}$

In the interval ${\displaystyle (0,\frac{3\pi }{4})}$, f(x) < 0 ⇒ curve is concave down.

In the interval ${\displaystyle (\frac{3\pi }{4},\frac{7\pi }{4})}$, f'(x) > 0 ⇒ curve is concave up.

In the interval ${\displaystyle (\frac{7\pi }{4},2\pi )}$, f'(x) < 0 ⇒ curve is concave down.

The curve is concave upward in ${\displaystyle (\frac{3\pi }{4},\frac{7\pi }{4})}$ and concave downward in ${\displaystyle (0,\frac{3\pi }{4})}$ and ${\displaystyle (\frac{7\pi }{4},2\pi )}$

f'(x) changes its sign when passing through x = ${\displaystyle \frac{3\pi }{4}}$ and x = ${\displaystyle \frac{7\pi }{4}}$

Now ${\displaystyle \text{f}\left(\frac{3\pi }{4}\right)=\sin \frac{3\pi }{4}+\cos \frac{3\pi }{4}}$

= ${\displaystyle \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}}$

= 0

${\displaystyle \text{f}\left(\frac{7\pi }{4}\right)=\sin \frac{7\pi }{4}+\cos \frac{7\pi }{4}}$

= ${\displaystyle -\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}}$

= 0

∴ The point of inflection are ${\displaystyle (\frac{3\pi }{4},0)}$ and ${\displaystyle (\frac{7\pi }{4},0)}$.