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Answer:
[0.25, 2]
Step-by-step explanation:
We have
4t² ≤ 9t-2
subtract 9t-2 from both sides to make this a quadratic
4t²-9t+2 ≤ 0
To solve this, we can solve for 4t²-9t+2=0 and do some guess and check to find which values result in the function being less than 0.
4t²-9t+2=0
We can see that -8 and -1 add up to -9, the coefficient of t, and 4 (the coefficient of t²) and 2 multiply to 8, which is also equal to -8 * -1. Therefore, we can write this as
4t²-8t-t+2=0
4t(t-2)-1(t-2)=0
(4t-1)(t-2)=0
Our zeros are thus t=2 and t = 1/4. Using these zeros, we can set up three zones: t < 1/4, 1/4<t<2, and t>2. We can take one random value from each of these zones and see if it fits the criteria of
4t²-9t+2 ≤ 0
For t<1/4, we can plug in 0. 4(0)²-9(0) + 2 = 2 >0 , so this is not correct
For 1/4<t<2, we can plug 1 in. 4(1)²-9(1) +2 = -3 <0, so this is correct
For t > 2, we can plug 5 in. 4(5)²-9(5) + 2 = 57 > 0, so this is not correct.
Therefore, for 4t^2 ≤ 9t-2 , which can also be written as 4t²-9t+2 ≤ 0, when t is between 1/4 and 2, the inequality is correct. Furthermore, as the sides are equal when t= 1/4 and t=2, this can be written as [0.25, 2]