using the interval [0,2pi]

and f(x) = sinx + cosx, obtain c £ (0,2pi) that satisfies the conclusion of Rolle's theorem
where £ mean element of and C means number
show step

1 answer

f(0) = f(2π) = 1
so the condition is satisfied.

So, now you want c such that f'(c)=0

f' = cosx - sinx
f' = 0 at x = π/4
π/4 is in the interval [0,2π], so ta-da!
Similar Questions
  1. Simplify #3:[cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] =
    1. answers icon 1 answer
  2. Solve this equation fo rx in the interval 0<=x<=3603sinxtanx=8 I would do it this way: sinxtanx = 8/3 sinx(sinx/cosx)=8/3
    1. answers icon 0 answers
    1. answers icon 3 answers
  3. tanx+secx=2cosx(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
    1. answers icon 0 answers
more similar questions