We have #f(x) = 2x^2 + 5x -3#, and we want to find the values of #x# for which #f(x) > 0#.
We start by finding the critical numbers.
Set #f(x) = 2x^2 + 5x -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 #-3# and #0.5#.
Now we check for positive and negative regions.
We have three regions to consider: (a) #x < -3#; (b) #-3 < x < 0.5#; and (c) #x > 0.5#.
Case (a): Let #x = -4#.
Then #f(4) = 2×(-4)^2 + 5(-4) -3 =32 – 20 -3 = 9#
#f(x)# is positive when #x < -3#.
Case (b): Let #x = 0#.
Then #f(0) = 2×0^2 + 5×0 -3 = 0 + 0 -3 = -3#
#f(x)# is negative when #-3 < x < 0.5#
Case (c): Let #x = 1#.
Then #f(1) = 2×1^2 + 5×1 -3 = 2 +5 – 3 = 4#
#f(x)# is positive when #x > 0.5#.
Answer: #f(x)# is positive when #x < -3# and #x > 0.5#.
