Find the absolute maximum value and the absolute minimum value of the followingfunctions in the given intervals:(i) `f(x)=x^2,x in [-2,2]` (ii) `f(x)=sinx+cosx ,x in [0,pi]`(iii) `f(x)=4x-1/2x^2, x in [-2,9/2]` (iv) `f(x)=(x-1)

Explanation

(i) `f(x) = x^(3), x in [-2, 2]`
` rArr f'(x) = 3x^(2) and f''(x) = 6x`
For maxima/minima
f'(x) = 0
` rArr 3x^(2) = 0`
` rArr x= 0 in [-2, 2]`
We find the values of f(x) at x = 0 and at the end points of the interval [-2, 2].
`:. f(0) = 0^(3) = 0`
` f(-2)=(-2)^(3)=-8`
` f(2) = 2^(3) = 8`
` :.` f(x) has absolute maximum value 8 at x = 2 and absolute minimum value- 8 at x = - 2.
(ii) Given function : `f(x) = sin x + cos x`
` rArr f'(x) = cos x- sin x `
For maxima/minima, f'(x) = 0
` rArr cos x - sin x = 0`
` rArr (sin x)/(cos x) = 1`
` rArr tan x = 1 rArr x = pi/4 in [0, pi]`
Now, we find the values of f(x) at point ` x = pi/4` and at the end points of the interval `[0, pi]`
at ` x= pi/4, f(pi/4) = sin pi/4 +cos pi/4 = 1/sqrt2+1/sqrt2 = sqrt2`
at ` x = 0, f(0) = sin 0 + cos 0 = 0 +1=1`
at ` x = pi, f(pi) = sin pi + cos pi = 0 - 1 =-1`
`:. f(x)" has absolute maximum value "sqrt2 at x = pi/4" and absolute minimum value "-1 at x = pi`.
(iii) Given function `f(x) = 4x - 1/2 x^(2)`
` rArr f'(x) = 4 - 1/2 (2x) = 4 - x`
For maxima/minima, f'(x) = 0
` rArr 4- x = 0 rArr x = 4 in [-2,9/2]`
Now, we find the values of f(x) at x = 4 and at the end point of the interval ` [-2, 9/2]`.
at ` x= 4, f(4) = 4(4) - 1/2 (4)^(2) =16 -8 = 8`
at ` x=-2, f(-2) = 4 (-2)-1/2 (-2)^(2)=-8-2=-10`
at ` x = 9/2, f(9/2) = 4 (9/2) - 1/2 (9/2)^(2)`
` = 18 - 81/8 = 63/8 = 7.875`
`:. ` f(x) has absolute maximum value 8 at x = 4 and absolute minimum value ` - 10 at x =- 2`.
(iv) Given function ` f(x) = (x-1)^(2) + 3`
` :. f'(x) = 2(x-1)`
For maxima/minima f'(x) = 0
` rArr 2(x-1)=0 rArr x = 1`
Now, we find the values of f(x) at the end points of the interval [-3, 1].
at ` x =1, f(1) =(1-1)^(2)+3=3`
at ` x=-3, f(-3) = (-3-1)^(2) + 3 = 19`
`:.` f(x) has absolute maximum value 19 at x =- 3 and absolute minimum value 3 at x = - 1.