Answer
Absolute maximum: $(2\pi,17.57)$
Absolute minimum: $(2.73,0.88)$
Work Step by Step
$ f(x)=2x+5\cos x \quad$ ... defined on $[0,2\pi].$
$ f^{\prime}(x)=2-5\sin x\quad$ ... defined on $(0,2\pi).$
$f^{\prime}(x)=0 \quad$ for
$2-5\sin x=0$
$\displaystyle \sin x=\frac{2}{5}$
$x=\displaystyle \sin^{-1}(\frac{2}{5})$
$x\approx 0.41,2.73\in(0,2\pi)\quad $(critical numbers)
$\left[\begin{array}{lccccccc}
x, \text{ interval} & 0 & (0,0.41) & 0.41 & (0.41,2.73) & 2.73 & (2.73,2\pi) & 2\pi\\
t=\text{ test number} & & 0.1 & & \pi/2 & & 3\pi/2 & \\
\text{ sign of }f^{\prime}(t) & & + & & - & & + & \\
f(x) & 5 & \nearrow & 5.41 & \searrow & 0.88 & \nearrow & 17.57 \\
& & & & & min & & max
\end{array}\right]$
Absolute maximum: $(2\pi,17.57)$
Absolute minimum: $(2.73,0.88)$
