#9:
The largest possible value is 5. This expression is maximized when x=3 and for any other value of x will cause the expression to be less than 5.
#10:
Since (x-4) is squared, this term is always ≥0. So whether (x-4)^2(x+3) is positive or negative totally depends on x+3.
x+3≥0 when x≥-3
The answer is [-3, infty)
#11:
We can start solving this by clearing the denominator.
(3-z)/(z+1) ≥ 1 can be written as 3-z ≥ 1(z+1)
Solving this, we have
3-z ≥ z+1
2 ≥ 2z
1 ≥ z
So, we start with z ≤ 1.
But, we have an issue that for some values of z less than 1 will cause us to end up with a negative value. That switch from positive values to negative will happen at z = -1, since the sign of the denominator will switch from positive to negative, positive if z > -1 and negative if z < -1.
So this leaves us with z > -1 AND z ≤ 1
The answer is (-1, 1].