# local Extrema, Absolute Extrema, antiderivatives

• Nov 18th 2009, 08:32 PM
Johnny Walker Black
local Extrema, Absolute Extrema, antiderivatives
1)State the intervals on which f(x)=cos(x) - sin(x) is increasing and decreasing on the interval [- , ] , and name all the local extrema.

(i don't know where to start on this problem.)

ANSWER: Increasing on [-∏ , -∏/4) and (3∏/4]; decreasing on (-∏/4 , 3∏/4).
Rel max (-∏/4 , √2) Rel min (3∏/4 , -√2).

2) Find the absolute extrema of f(x)=sin(x) - cos(x) on the interval [0,∏]
f'(x)=-cos(x) + sin(x)
g(0)=0
g(∏)=1
This is where i'm stuck.

ANSWER: Abs max √2 at x=3∏/4

3) Use the given information to find f. f'(x)= 3x² -x + 4 f(1)=2
f(x)= x³ -½x² + 4x + c
1 + -½ + 4 + c and f(1) = 2
This is where i stopped. How did my professor get f(x) = x³ -½x² + 4x -5/2 ??

ANSWER: f(x) = x³ -½x² + 4x - ⁵⁄₂

Need Help! Thanks everyone!
• Nov 18th 2009, 11:22 PM
earboth
Quote:

Originally Posted by Johnny Walker Black
...
3) Use the given information to find f. f'(x)= 3x² -x + 4 f(1)=2
f(x)= x³ -½x² + 4x + c
1 + -½ + 4 + c and f(1) = 2
This is where i stopped. How did my professor get f(x) = x³ -½x² + 4x -5/2 ??

ANSWER: f(x) = x³ -½x² + 4x - ⁵⁄₂

From

$\displaystyle f(x)=x^3-\frac12 x^2+4x+c$ you'll get

$\displaystyle f(1)=2=1-\frac12+4+c~\implies~2=5-\frac12+c~\implies~2-5+\frac12 = c$

and you'll see that your prof is right.