Question

Locate the absolute extrema of the function f(x) = sin() on the closed interval 0, [0.5] Select one: a. The absolute minimum is 0, and it occurs at the left endpoint x=0. 13 The absolute maximum is and it occurs at the right endpoint X=- 3: O b. The absolute minimum is 0, and it occurs at the right endpoint x = 3: 1 The absolute maximum is 2, and it occurs at the left endpoint x=0. c. The absolute minimum is 0, and it occurs at the left endpoint x=0. The absolute maximum is 7, and it occurs at the right endpoint 1 X=- 3. = tý d. The absolute minimum is 0, and it occurs at the right O

Answer 1

-- راجا d sin (1) fen) XE [onth Guiven, f(x)= sin (1x) XE [o. 1 / 3 Now, differentiate eqn is with respect to ce, we get f'(x) dre cos o 7,0 for π cos (Tlu) 3 cos (TTX) 70 for SO, #lo) T cos (o)= T 70 and xo Therefore, for at To 1/3 fixes is increasing function for x € [o. +3] "It is also clear from the graph of fcu) = sin(x) that fexs is increasing for XE [0, I/]. f (1/3) - IT cos ( 13 ) = 17 f'(x) to 1 Mu 21 1 - 1 Ed -1 occurs at L=0 . The absolute minimum ie, f10)=sin(o)=0 and, maximum occurs at I abrolute je f ($) = sin ( ) = 2