Find the largest value of function y=cos²x
Answers
Answer:
1 becoz the values of cos will be greater than -1 and less than 1 there fore 1 is the answer
Answer:
y=cos(2x)-x ==>y'= -2sin(2x)-1 and y''= -4cos(2x). Equating y' to 0, gives sin(2x)=(-1/2) =sin(-π/6) ==> 2x=nπ+[(-1)^n](-π/6) ==>x=(nπ/2)+[(-1)^(n+1)](π/12). Hence the values lying in [-π/2,π/2] are x = -π/12, -5π/12, ==> 2x = -π/6, -5π/6 ==> y”=(-4cos(2x)) = -2√3, 2√3 at -π/12 and -5π/12, respectively. Hence -π/12 is a local maximum point and -5π/12 is a local minimum point.
Now y(-π/12) = cos(-π/6)+π/12=(√3)/2+π/12 ~1.128. and y(-5π/12)=cos(-5π/6) - (-5π/12) = (5π/12) - (√3)/2 ~ 0.443. Also the boundary points of the given interval should be considered:
y(-π/2) = cos(-π)-(-π/2) = -1+1.5708 = 0.5708 and y(π/2) = cos(π)-(π/2) = (-1) - 1.5708 = -2.5708. Comparing the values of y at the local maximum point (-π/12) and the end-points -π/2 and π/2, we conclude that the absolute maximum value of y occurs at -π/12 and the maximum value of y is nearly 1.128.
For the smallest value, we see that the value of y at the only local minimum point (-5π/12) is ~0.443, whereas at the end-point π/2 the value of y is -(π/2)-1 ~ (-2.5708).
Thus the minimum value of y is nearly (-2.5708) and the maximum value is nearly 1.128