Answer
$f(\frac{\pi}{8}) = 0.25$ m
$f'(\frac{\pi}{8}) = 4$ m s$^{-1}$
Work Step by Step
Let $y = f(x)$
1. Find $f(\frac{\pi}{8})$
$f(\frac{\pi}{8}) = \frac{1}{3} cos(12 \times \frac{\pi}{8}) - \frac{1}{4} sin ( 12 \times \frac{\pi}{8})$
$f(\frac{\pi}{8}) = \frac{1}{3} cos(\frac{12\pi}{8}) - \frac{1}{4} sin (\frac{12\pi}{8})$
$f(\frac{\pi}{8}) = (\frac{1}{3} \times cos(\frac{3\pi}{2})) - (\frac{1}{4} \times sin (\frac{3\pi}{2}))$
$f(\frac{\pi}{8}) = (\frac{1}{3} \times 0) - (\frac{1}{4} \times (-1))$
$f(\frac{\pi}{8}) = - (-\frac{1}{4})$
$f(\frac{\pi}{8}) = \frac{1}{4}$
$f(\frac{\pi}{8}) = 0.25$ m
2. Find $f'(\frac{\pi}{8})$
$f'(x) = [(\frac{1}{3})(-sin(12x))(12)] - [(\frac{1}{4})(cos(12x))(12)]$
$f'(x) = [(\frac{12}{3})(-sin(12x))] - [(\frac{12}{4})(cos(12x))]$
$f'(x) = [4(-sin(12x))] - [3(cos(12x))]$
$f'(x) = -4sin(12x) - 3cos(12x)$
$f'(\frac{\pi}{8}) = -4sin(12 \times \frac{\pi}{8}) - 3cos(12 \times \frac{\pi}{8})$
$f'(\frac{\pi}{8}) = -4sin(\frac{3\pi}{2}) - 3cos(\frac{3\pi}{2})$
$f'(\frac{\pi}{8}) = (-4 \times -1) - (3 \times 0)$
$f'(\frac{\pi}{8}) = (4) - (0)$
$f'(\frac{\pi}{8}) = 4$ m s$^{-1}$
