Question

The derivative of the function f(0) = 96x + 12x2 – 20x3 + 3x4 has three roots:-1, 2, and 4. That is, f'(-1) = f'(2) = f'(4) = 0 and there are no other points where f'(x) = 0. Find the maximum and the minimum value of f on the interval (0,3). Explain how you arrive at your answer.

Answer 1

f(0) = 9676 +120€ -2003 + 384 now has 3 roots ,-1,2 andy Critical points are solution of f'(o)=0 -> ale -1,2,4 Critical points we have to find maximum and minimum value of f(x) in on the on the interval [0,3] Critical points 26{0,3] check By second derivative test the let us N=2 have manimum or minimum firma d 0962 +12*4-2023 +304) da GON² + 12x² flrada 96 + 242 60% +12.023) d ( 96 +248 da 24 -12034 + 12x376? 24 – 12094 +302= 12(2-106 +33e?) fin) - So at n=2 f(2) = 12( 2-10(2) +362)2) = 12'/ 8-20+12) = 12 * (-6) Lo so f(a) has manimum at a = 2

now and let us checked at end points of 10,3] 9600) +1260- 2010)] +360)4 و (د) f 13 ) 90x3 + 12 (3)? - 2013)? +363)4 99 + and at : 2 f2) 90(2) + 1212)?-2012) +2614 126 .f()=0 f(3): 99 f(2) =128 on [Elo] of(x) has Whiletrobue 128 at x2 and minimum at 2:0