The LHS can be broken into parts:
(x-3)(x-5) on (-∞,3] or [5,∞)
-(x-3)(x-5) on (3,5)
The RHS similarly
x(x-4) on (-∞,0] or [4,∞)
-x(x-4) on (0,4)
This gives 5 regions
(1): (∞,0]
x^2-8x+15 <= x^2-4x
x >= 15/4 not in range
(2): (0,3)
x^2-8x+15 <= -x^2+4x
(6-sqrt6)/2 <= x <= (6+sqrt6)/2
in range then (6-sqrt6)/2 <= x < 3
(3): [3,4)
-x^2+8x-15 <= -x^2+4x
x <= 15/4
in range then 3 < x <= 15/4
(4): [4,5)
-x^2+8x-15 <= x^2-4x
x <= (6-sqrt(6))/2 or x>= (6+sqrt6)/2
in range then (6+sqrt6)/2 >= x >5
(5): [5,∞)
Same as (1)
but now all in range: x>=5
Put this all together for the Final answer:
(6-sqrt6)/2 <= x <= 15/4 or x >= (6+sqrt6)/2
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Posted by Jer
on 2024-09-20 14:03:49 |
Solution