# Maximum Value of a function on a closed interval

• Apr 12th 2010, 08:14 PM
Manabu
Maximum 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 PM
pickslides
Quote:

Originally Posted by Manabu
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?
• Apr 12th 2010, 08:27 PM
Debsta
Quote:

Originally Posted by Manabu
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.

A couple of suggestions:
Have you learnt to write that function in the form sin (a+b) or cos(a+b)?
If so, that'd be the easiest way.
2. Calculus??
• Apr 12th 2010, 08:28 PM
Manabu
Quote:

Originally Posted by pickslides
$\displaystyle f(x)=4\sin x-3\cos x$

$\displaystyle f'(x)=4\cos x+3\sin x$

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?

Not really. I understand how you got to that point, but I'm not 100% sure how to calculate maximum values.
• Apr 12th 2010, 08:36 PM
Failure
Quote:

Originally Posted by Manabu
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.

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.