Question

f(0.5) = 60, meaning that 0.5 seconds after the object was dropped, the object was dropped feet above the ground. this interpretation makes sense in the context of the problem. f(7) = -720, meaning that 7 seconds after the object was dropped, the object was above feet above the ground. this interpretation does not make sense in the context of the problem. based on the observations above, it is clear that an appropriate domain for the function is all real numbers non-negative real numbers real numbers in a≤x≤b integers whole numbers (0, 1, 2, ...) positive integers (1, 2, 3, ...) integers in a≤x≤b

Explanation:

Step1: Analyze valid time values

Time is a continuous quantity here, and we only consider times from when the object is dropped ($t=0$) until it hits the ground (when height = 0). We first find when $f(t)=0$:
Assuming the height function is $f(t) = -16t^2 + 80$ (since $f(0.5)=76$: $-16(0.5)^2 +80 = -4 +80=76$, and $f(7)=-16(49)+80=-784+80=-704$ which is close to the given -720, likely a typo, but the key is negative height is invalid). Solve $-16t^2 +80=0$:

$$\begin{align*} 16t^2&=80\\ t^2&=5\\ t&=\sqrt{5}\approx2.24 \end{align*}$$

So valid $t$ is between 0 and $\sqrt{5}$.

Step2: Match to domain type

Valid times are non-negative real numbers within a closed interval $0\leq t\leq\sqrt{5}$.

• $f(0.5)=76$: 0.5 seconds after release, the object is 76 feet above the ground, which is valid as time can be a non-integer and positive height makes sense.
• $f(7)=-720$: 7 seconds after release, the object would be -720 feet above the ground, which is impossible (a negative height has no physical meaning here, as the object would have hit the ground much earlier).
• The domain must include all valid non-negative real times from release until the object hits the ground, which corresponds to real numbers in a closed interval $a\leq x\leq b$.

Answer:

real numbers in $a\leq x\leq b$