Question

6.
(a)
Find the absolute maximum value and the absolute minimum value of
1/3x^3– 3 x² + 5 x + 8 in [0, 4].
​

Answer 1

for absolute Max or Min....

f'(x)=0 is necessary condition....

so f'(x)=x²-6x+5 must be 0..

so x²-6x+5=0...

by solving it we get x=5 or x=1...

there are two values. in which one held Max and other held Min....

for that we will find f"(x)....

so f"(x)=2x-6...

now,f"(5)=10-6=4>0 it means at x=5 there is minimum value....which is -1/3

now,f"(1)=2-6=-4<0 it means at x=1 there is maximum value....which is 31/3

Thanks.....