+9491 At 1 second and at 3 seconds is correct for c .
(Notice how on this graph, when t = 1 and t = 3, the tangent line is totally "flat" with a slope of zero.)
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Alice is moving backwards when the velocity is negative.
Here is a graph of the velocity function for reference: https://www.desmos.com/calculator/dqfc3ssois
(Note that in the graphs, x represents t )
We know that the graph is continuous everywhere,
AND we know that the only place where the graph touches the horizontal axis is at t = 1 and at t = 3
So is the graph above or below the horizontal axis when t < 1 ?
To find that out, we can test any value that is less than 1 . So let's use t = 0
v(0) = 3(0)2 - 12(0) + 9
v(0) = 9
9 is positive, so when t = 0 , the velocity is positive.
And we can conclude that, for ALL values of t less than 1 , the velocity is positive.
v(t) > 0 for all values of t in the interval (-∞, 1)
Now we want to know is the graph above or below the horizontal axis when t is between 1 and 3 ?
To find that out, we can test any value that is between 1 and 3 . So let's use t = 2
v(2) = 3(2)2 - 12(2) + 9
v(2) = -3
-3 is negative, so when t = 2 , the velocity is negative.
And we can conclude that, for ALL values of t between 1 and 3 , the velocity is negative.
v(t) < 0 for all values of t in the interval (1, 3)
Now we want to know is the graph above or below the horizontal axis when t is greater than 3 ?
To find that out, we can test any value that is greater than 3 . So let's use t = 4
v(4) = 3(4)2 - 12(4) + 9
v(4) = 9
9 is positive, so when t = 4 , the velocity is positive.
And we can conclude that, for ALL values of t greater than 3 , the velocity is positive.
v(t) > 0 for all values of t in the interval (3, ∞)
So now we know that Alice is moving backwards when t is in the interval (1, 3)
(Notice how on this graph, when t is in the interval (1, 3), the tangent line is "downhill" with a negative slope.)
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