On the closed interval [0,2pi], the maximum value of the function f(x)=4sinx-3cosx is:
A.)3
B.)4
C.)24/5
D.)5
E.)None of these
Explanations appreciated.
$\displaystyle f(x)=4\sin x-3\cos x $
$\displaystyle f'(x)=4\cos x+3\sin x $
for max let $\displaystyle f'(x)=0 $
$\displaystyle 0=4\cos x+3\sin x $
$\displaystyle 4\cos x=-3\sin x $
$\displaystyle \frac{-4}{3}=\frac{\sin x}{\cos x} $
$\displaystyle \tan x =\frac{-4}{3} $
Can you finish this?
D. is the correct answer, and it can be found without calculus (but with some trigonometry), because
$\displaystyle 4\sin x-3\cos x=5\cdot\left(\frac{4}{5}\cdot \sin x-\frac{3}{5}\cdot\cos x\right)=-5\cos(x+\varphi)$
for $\displaystyle \varphi := \arccos(3/5)$. Now, since x varies over a whole period of the function, the maximum value of f(x) must be 5.