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#.
If #2x^2 – 7x + 3 ≤ 0# when #0.5 ≤ x ≤ 3#, then
#2x^2 – 7x ≤ -3# when #0.5 ≤ x ≤ 3#
