Task 1: abs(7-3x) + 4x > 5
Task 2: sqrt(7-3x) <= x -1
Task 3: sqrt(5-2x) > 1 - x
Task 4: x^2*(1 + 4x^2) < abs(x - 2x^2) + 6 + 4x^3
If 7- 3x< 0 (that is, if x> 7/3) this is 3x- 7+ 4x> 5. Can you solve that? If so that , this is . Can you solve that?
is positive and if a< b, then so this is the same as [itex]7- 3x\le (x- 1)^2[/itex]. Can you solve that?Task 2: sqrt(7-3x) <= x -1
Same thing as task 2.]Task 3: sqrt(5-2x) > 1 - x
As in task 1, consider the cases and separately. The equation has solutions x= 0 and x= 1/2. If x= -1, so for all x < 0. If x= 1/4, then so for all x between 0 and 1/2. Finally, if x= 2, so that for all x> 1.Task 4: x^2*(1 + 4x^2) < abs(x - 2x^2) + 6 + 4x^3
That is, there are two cases:
1) x< 0 or x> 1/2. In this case, so and the inequality becomes which is the same as which is itself the same as . The simplest way to solve that is to find the solutions to and check points between those solutions to see where they are "> 0" and where "< 0".
2) 0< x< 1/2. In this case, so and the inequality becomes which is the same as which is itself the same as .