Question

need help on attachment

Answer 1

the answer is not number not number 3 or 1 because I already tried those and got it wrong

Answer 2

f(x) = (1/3)x^3 -5x^2 + 16x + 9
f'(x) = x^2 - 10x + 16
f'(x) > 0 when x^2 - 10x + 16 > 0
(x - 5)^2 - 25 + 16 > 0
(x - 5)^2 > 9
|x - 5| > 3
or
3 < |x - 5|
or
3 < x - 5 < -3
3 + 5 < x < -3 + 5
8 < x < 2

so f(x) is increasing when in (-inf,8) and (2,inf)

f(x) <= 0 when 2 <= x <= 8

f(x) = 0 when (x - 5)^2 - 25 + 16 = 0
or when |x - 5| =3
or when x - 5 = 3 or x - 5 = -3
x - 5 = 3 => x = 8
x - 5 = -3 => x = 2

f''(x) =2x - 10
f''(8) = 2*8 - 10 = 6
f"(2) = -6

since f''(2) < 0 and f'(2) = 0, f(2) is a local maximum
since f'(8) > 0 and f'(8) = 0, f(8) is a local minimum
but since f'(x) > 0 when 8 < x < 2, they are not global extrema

in [0,3] f(x) has a maximum at f(2) but f(8) is not a minimum in this region
since f'(x) > 0 in (-inf,2) and f(2) is a maximum, f(0) < f(2)

Hence, f(0) < f(2) and f(3) < f(2), the minimum in [0,3] is either f(0) or f(3)

f(0) = 9
f(3) = 9 - 45+ 48 + 9
= 18 + 3
= 21

Since f(0) < f(3), f(0) = 9 is the absolute minimum

Ans. 5) absolute minimum is 9.