Question -
Answer -
(i) The given function is f(x)= x3.
Then, we evaluate the value of f atcritical point x = 0 and at end points of the interval [−2,2].
f(0) = 0
f(−2) = (−2)3 =−8
f(2) = (2)3 =8
Hence, we can conclude that the absolutemaximum value of f on [−2, 2] is 8 occurring at x =2. Also, the absolute minimum value of f on [−2, 2] is −8occurring at x = −2.
(ii) The given function is f(x)= sin x + cos x.
Then, we evaluate the valueof f at critical point
and at the end points of the interval[0, π].
Hence, we can conclude thatthe absolute maximum value of f on [0, π] is
occurring atand the absolute minimum value of f on[0, π] is −1 occurring at x = π.
(iii) The given function is
Then, we evaluate the valueof f at critical point x = 4 and at the endpoints of the interval
.
Hence, we can conclude thatthe absolute maximum value of f onis 8 occurring at x =4 and the absolute minimum value of f on is −10 occurring at x = −2. (iv) The given function is
.
Now,
2(x − 1) = 0 ⇒ x = 1
Then, we evaluate the value of f atcritical point x = 1 and at the end points of the interval[−3, 1].
Hence, we can conclude that the absolutemaximum value of f on [−3, 1] is 19 occurring at x =−3 and the minimum value of f on [−3, 1] is 3 occurringat x = 1.