Question

The function f is defined by f(x) = 3x - 4cos(2x + 1), and its derivative is f'(x) = 3 + 8sin(2x + 1). What are all values of that satisfy the conclusion of the Mean Value Theorem applied to f on the interval [-1, 2] ?

Answer 1

Answer:

f ( x )  does not satisfy the mean value theorem

Step-by-step explanation:

The given data :-

• f(x) = 3x - 4 cos ( 2x + 1 )
• f'(x) = 3 +8 sin ( 2x + 1 )
• The interval [ -1 , 2 ]

Solution :-

i) f(x) is continuous on [ -1 , 2 ]

ii) f(x) is derivable on ( -1 , 2 )

f ( -1 ) = 3 * (-1 ) - 4 cos [ 2 * ( -1 )  + 1 ]  = - 3 + cos (-1 ) = - 3 - 4 * 0.9998 = - 6.992

f ( 2 ) = 3 * 2 - 4 cos ( 4 * 2 + 1 ) = 6 + 4 cos 9  = 6 - 3.9507 = - 3.04924

iii) f ( -1 ) ≠ f ( 2 )

f ( x ) is not real valued function so it does not satisfy the mean value theorem