Question -
Answer -
Let┬аf(x) = 2x3┬атИТ24x┬а+ 107.
We first consider the interval [1, 3].
Then, we evaluate the value of┬аf┬аatthe critical point┬аx┬а= 2 тИИ [1, 3] andat the end points of the interval [1, 3].
f(2) = 2(8)тИТ 24(2) + 107 = 16 тИТ 48 + 107 = 75
f(1) = 2(1)тИТ 24(1) + 107 = 2 тИТ 24 + 107 = 85
f(3) = 2(27)тИТ 24(3) + 107 = 54 тИТ 72 + 107 = 89
Hence, the absolute maximum value of┬аf(x)in the interval [1, 3] is 89 occurring at┬аx┬а= 3.
Next, we consider the interval [тИТ3, тИТ1].
Evaluate the value of┬аf┬аatthe critical point┬аx┬а= тИТ2 тИИ [тИТ3, тИТ1]and at the end points of the interval [1, 3].
f(тИТ3) = 2(тИТ27) тИТ 24(тИТ3) + 107 = тИТ54 + 72 + 107 = 125
f(тИТ1) =2(тИТ1) тИТ 24 (тИТ1) + 107 = тИТ2 + 24 + 107 = 129
f(тИТ2) =2(тИТ8) тИТ 24 (тИТ2) + 107 = тИТ16 + 48 + 107 = 139
Hence, the absolute maximum value of┬аf(x)in the interval [тИТ3, тИТ1] is 139 occurring at┬аx┬а= тИТ2.