SOLUTION: Given a>b>c>0, the function f(x)= x(x-a)(x+b)(x-c) is less than or equal to zero on the intervals 9 a. (-∞, -b), [0,a] and [c,∞) b (-∞,-b], [0,c] and [a,∞) c. [-b,0] a

Algebra ->  Finance -> SOLUTION: Given a>b>c>0, the function f(x)= x(x-a)(x+b)(x-c) is less than or equal to zero on the intervals 9 a. (-∞, -b), [0,a] and [c,∞) b (-∞,-b], [0,c] and [a,∞) c. [-b,0] a      Log On


   

Question 1181602: Given a>b>c>0, the function f(x)= x(x-a)(x+b)(x-c) is less than or equal to zero on the intervals
9
a. (-∞, -b), [0,a] and [c,∞)
b (-∞,-b], [0,c] and [a,∞)
c. [-b,0] and [c,a]
d.[-a,-c] and [0,b]
Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!

x%28x-a%29%28x%2Bb%29%28x-c%29+%3C=0, if a%3Eb%3Ec%3E0


=> all solutions
x+%3C=0
x+%3C=a
x%3C=-b
x+%3C=c
so, all combinations are
a%3E0,
0%3Cb%3Ca,
0%3Cc%3Cb,
-b%3C=x%3C=0
or
a%3E0,
0%3Cb%3Ca,
0%3Cc%3Cb,
c%3C=x%3C=a

Interval notation:
(0, infinity)
(0, a)
(0, b)
[-b,+0]
(0, infinity)
[c,+a]


your option: if a%3Eb%3Ec%3E0
c. [-b,0] and [c,a]