Finding the extrema of $f(x) = \sin x + \cos x$

Question

Find all extrema in the interval $[0, 2\pi]$ for $y=\sin(x) + \cos(x)$

I got MAX:$(0,1), (2\pi, 1)$ / MINI:$(5\pi/4, -\sqrt{2})$

I am not sure about the max points.

Answer 1

Your minimum is correct. You must have had a computational error in the maximum or something - but notice that $y(x+\pi)=-y(x)$, which implies that if $\frac{5\pi}4$ is a minimum, then that minus $\pi$ - at $\frac{\pi}4$ must be a maximum (at least locally; you would still want to check it against the endpoints, but it has value $\sqrt{2}$ there, so is greater)

Answer 2

\begin{align*} f(x) & = \cos x + \sin x \\ \Rightarrow f'(x) & = - \sin x + \cos x \\ \Rightarrow f''(x) & = - \cos x - \sin x \end{align*}

So we need $ x \in [0, 2 \pi )$ for which $f'(x) = 0$. Finally, if $ f''(x) < 0$, then $x$ is a local maximum, and likewise if $f''(x) > 0$, then $x$ is a local minimum.

Answer 3

To find the critical points of $f(x) = \sin x + \cos x$ in the interval $[0, 2\pi]$, set $f'(x) = 0$.

$$f'(x) = \cos x - \sin x = 0 \Rightarrow \cos x = \sin x \Rightarrow 1 = \tan x$$

Hence, the critical points occur at $x = \dfrac{\pi}{4}$ and $x = \dfrac{5\pi}{4}$.

The line analysis shows that the derivative changes from positive to negative at $x = \dfrac{\pi}{4}$ and from negative to positive at $x = \dfrac{5\pi}{4}$. Thus, by the First Derivative Test, the function has a relative maximum at $x = \dfrac{\pi}{4}$ and a relative minimum at $x = \dfrac{5\pi}{4}$.

The absolute extrema could occur at the endpoints of the closed interval. However,

$$f(0) = f(2\pi) = 1 < \sqrt{2} = f\left(\frac{\pi}{4}\right)$$

so the function has an absolute maximum at $x = \dfrac{\pi}{4}$ and

$$f(0) = f(2\pi) = 1 > -\sqrt{2} = f\left(\frac{5\pi}{4}\right)$$

so the function has an absolute minimum at $x = \dfrac{5\pi}{4}$, as you concluded correctly.