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.

Printable View

- Apr 12th 2010, 08:14 PMManabuMaximum Value of a function on a closed interval
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. - Apr 12th 2010, 08:20 PMpickslides
$\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? - Apr 12th 2010, 08:27 PMDebsta
- Apr 12th 2010, 08:28 PMManabu
- Apr 12th 2010, 08:36 PMFailure
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.