calculus...?

Question

verify the conditions of the Rolle's theorem. if the conditions are satisfied, find c.

1. f(x) = sin2x; [0, 1/2π]

2. f(x) = x^3 - 2x- x + 2; [1,2]

Answer 1

1) Conditions:
sin(2*0) = sin(0) = 0
sin(2 * pi/2) = sin(pi) = 0

c:
2cos(2c) = 0
cos(2c) = 0
2c = pi/2
c = pi/4

2) conditions:
f(1) = 1 -2 -1 +2 = 0
f(2) = 8 - 4 - 2 + 2 = 4

As you've typed the formula, Rolle's theorem is not applicable. Are you sure you typed it correctly?

Edit: You might have meant x^3 - 2x^2 -x +2, in which case...

conditions:

f(1) = 1 - 2 - 1 + 2 = 0
f(2) = 8 - 8 - 2 + 2 = 0

And the conditions are met.

f'(c) = 3c^2 - 4c - 1 = 0
use quadratic formula to find c = (2 + sqrt(7)) / 3

Answer 2

sin0 = sinpi=0
The function is cont.and deriv,
2cos2x=0 so cos 2x= 0 2x = pi/2 (in the given interval and x=pi/4.
2)The same cond are verified f(1)=-0 =f(2)
3x^2-4x-1=0
x=((4+sqrt(16+12))/6 =2/3+1/3 sqrt(7)=1.5486

Answer 3

I pity myself for having chosen to open Answers

Answer 4

all you people with math and physics questions just looking for people to do your work makes me sick. Appearently now the internet is just another source for cheating.