Since the function is h(t) = −5t² + 15t, the domain of the function h(t) is 0 ≤ t ≤ 3
The domain of a function is the range of input values that make the function have real values.
Now given the function h(t) = −5t² + 15t, it will have real values when
h(t) ≥ 0
So, h(t) ≥ 0
⇒ −5t² + 15t ≥ 0
Factorizing, we have
-5t(t - 3) ≥ 0
5t(t - 3) ≤ 0
t(t - 3) ≤ 0
t ≤ 0 and t - 3 ≤ 0
t ≤ 0 and t ≤ 3
For t ≤ 0, say -1, t(t - 3) = -1(-1 - 3) = -1(-4) = 4 ≥ 0
For 0 ≤ t ≤ 3, say 2, t(t - 3) = 2(2 - 3) = 2(-1) = -2 ≤ 0
For t ≥ 3, say 4, t(t - 3) = 4(4 - 3) = 4(1) = 4 ≥ 0
Since we require t(t - 3) ≤ 0, the domain of t is thus 0 ≤ t ≤ 3
So, the domain of the function h(t) is 0 ≤ t ≤ 3
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