# Graphing using 1st and 2nd derivatives

• November 1st 2010, 06:06 PM
looseenz2
Graphing using 1st and 2nd derivatives
f(x) = 3 sin x + 3 cos x

0 ≤ x ≤ 2π

How do I use this to find local max & min, critical points, and inflection points?
• November 2nd 2010, 02:52 AM
Unknown008
I'm not sure of what you have to do, but if you have to graph f(x), you can convert it into a simple form:

$3\sin x + 3 \cos x = R \sin(x + \alpha)$

$R = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18}$

$\alpha = \tan^{-1}\left(\dfrac33\right) = \dfrac{\pi}{4}$

$3\sin x + 3 \cos x = \sqrt{18} \sin\left(x + \dfrac{\pi}{4}\right)$

The first and second derivative should also be easier to graph now, provided you know how to graph f(x).
• November 2nd 2010, 04:08 AM
HallsofIvy
Quote:

Originally Posted by looseenz2
f(x) = 3 sin x + 3 cos x

0 ≤ x ≤ 2π

How do I use this to find local max & min, critical points, and inflection points?

"Critical points" are, by definition, points where the first derivative is 0. "Local max and min" are points where the first derivative changes sign (and so the second derivative is not 0), and inflection points are where the second derivative is 0.

f'(x)= 3cos(x)- 3 sin(x)

That will equal 0 when cos(x)- sin(x)= 0 or cos(x)= sin(x). That happens when $x= \pi/4$ or $x= 5\pi/4$.

f''(x)= -3 sin(x)- 3cos(x)

That will equal 0 when sin(x)+ cos(x)= 0 or cos(x)=-sin(x). That happens when $x= 3\pi/4$ or $x= 7\pi/4$.

( $\sqrt{3^2+ 3^2}= \sqrt{2(3^2)}= 2\sqrt{3}$, not $\sqrt{19}$.)
• November 2nd 2010, 04:20 AM
Unknown008
Oops... dang (Doh)