Let's modify the inequality by adding 3 to each side. Then
2x^2 – 7x + 3 ≤ 0
We start by finding the critical numbers.
Set f(x) = 2x^2 – 7x + 3 = 0 and solve for x.
(2x-1)(x - 3) = 0
2x-1 = 0 or x-3 = 0
2x = 1
x = ½ = 0.5 or x = 3
The critical numbers are 0.5 and 3.
Now we check for positive and negative regions.
We have three regions to consider: (a) x ≤ 0.5; (b) -3 ≤ x ≤ 0.5; and (c) x ≥ 0.5.
Case (a): Let x = 0.
Then f(0) = 2×0^2 - 7×0 + 3 =0 - 0 + 3 = 3
f(x) ≥ 0 when x ≤ 0.5.
Case (b): Let x = 1.
Then f(1) = 2×1^2 - 7×1 + 3 = 2 - 7 + 3 = -2
f(x) ≤ 0 when 0.5 ≤ x ≤ 3
Case (c): Let x = 4.
Then f(1) = 2×4^2 - 7×4 + 3 = 32 - 28 + 3 = 7
f(x) ≥ 0 when x ≥ 3.
![@@@Sign Chart]()
If 2x^2 – 7x + 3 ≤ 0 when 0.5 ≤ x ≤ 3, then
2x^2 – 7x ≤ -3 when 0.5 ≤ x ≤ 3
