Asked by Ryu
h(t)=-t^2+6f+7 modes height (ft) at time t(sec) what is the average rate of change on [1,5] and what does it mean in context
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GPT-5 mini
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I assume the function is h(t) = −t^2 + 6t + 7 (the "6f" looks like a typo for 6t). h(t) is height in feet and t is time in seconds.
Average rate of change on [1,5]:
(h(5) − h(1)) / (5 − 1).
Compute:
h(5) = −25 + 30 + 7 = 12
h(1) = −1 + 6 + 7 = 12
So average rate = (12 − 12) / 4 = 0 ft/sec.
Meaning in context:
Over the time interval from 1 s to 5 s the net change in height is zero — the object is at the same height at t = 1 and t = 5. On average the height did not change (0 ft/sec), but that hides motion in between: the height rises to a maximum at t = 3 (h(3) = 16 ft) and then falls, so the increase and decrease cancel out over [1,5]. (Instantaneous rates: h′(t) = −2t + 6, so h′(1)=4 ft/s upward and h′(5)=−4 ft/s downward.)
Average rate of change on [1,5]:
(h(5) − h(1)) / (5 − 1).
Compute:
h(5) = −25 + 30 + 7 = 12
h(1) = −1 + 6 + 7 = 12
So average rate = (12 − 12) / 4 = 0 ft/sec.
Meaning in context:
Over the time interval from 1 s to 5 s the net change in height is zero — the object is at the same height at t = 1 and t = 5. On average the height did not change (0 ft/sec), but that hides motion in between: the height rises to a maximum at t = 3 (h(3) = 16 ft) and then falls, so the increase and decrease cancel out over [1,5]. (Instantaneous rates: h′(t) = −2t + 6, so h′(1)=4 ft/s upward and h′(5)=−4 ft/s downward.)
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