Answer
(a) The maximum value is approximately $-1.15$
The minimum value is approximately $-2.25$
(b) The maximum value is $-1.17$
The minimum value is $-2.26$
Work Step by Step
(a) $f(x) = x-2~cos~x,~~~-2 \leq x \leq 0$
We can use the zoom function on a graphing calculator to estimate the maximum and minimum values of the function in the interval $[-2,0]$
The maximum value is approximately $-1.15$
The minimum value is approximately $-2.25$
(b) $f(x) = x-2~cos~x$
$f'(x) = 1+2~sin~x = 0$
$sin~x = -\frac{1}{2}$
$x = arcsin(-\frac{1}{2})$
$x = -\frac{\pi}{6}$
When $x = -\frac{\pi}{6}$:
$f(-\frac{\pi}{6}) = (-\frac{\pi}{6})-2~cos~(-\frac{\pi}{6}) = -2.26$
The minimum value is $-2.26$
When $x = -2$:
$f(-2) = (-2)-2~cos~(-2) = -1.17$
The maximum value is $-1.17$
