Question

12 consider the function f(x) = (1-x²) 1/3 over the interval [0,3]. a) The absolute maximum value of f over the interval is at x = B) The absolute minimum value off over the interval is at x=

Answer 1

f(w) = (1-22) au =O .. x=0 f(1) = (1-12343 Solutioni given that for critical points fo= f'(n) *(1-x217%. (0-2)») 3(1- 22/2/ f'(n) 311 - 12)-2/3 az il But for the interval given [0, 3] and n & -1 out for the Intervals. 80 for the intervale [0, 3] f(0) = (- 02% f(is) = (1-152) 70 772 17 = -1.077217 f (2.0) = (1-2.0² ) Vz -1.44224 f (2-5) = (1-2.52) V3 - 1.73 8013322 f (3-6) = (1 - 3:02) -2 x = 0 and x 7 - ا رر)

.: over the Internal [0 A) The absolute manimum value off at n=0 f(0) = (1 - 021/3 B) The absolute minimum value of af over the Interval (1,3) or (1,3 at n= 3 f(3) = (1-32% le Answer