Does anyone know how to do this there is another page but if you can’t see it zoom in
1 answer:
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Answer:
Yes, at a time t such that (√2)/2 ≤ t ≤ 2.
Explanation:
To answer the question
Therefore, where the domain of the function is the set of all real numbers x for which f(x) is a real number we have
For Chloe's velocity
Finding the boundaries of the function gives;
and
At t = 1, we have
We find the maximum point as follows;
From which we have;
or
∴ 1 = 2·t² and from which t = (√2)/2
Hence the function C(x) is decreasing from t = (√2)/2 to t = 2
For Brandon
For 0 ≤ t ≤ 1, 1 ≤ B(t) ≤8 and for 1 < t ≤ 2, 8 < B(t) ≤ 1.5
1 ≤ f(x) ≤ 1.5
Given that the function B(t) is differentiable, therefore, continuous, there exists a point at which the function C(t) and B(t) intersects given that;
For 0 ≤ t ≤ (√2)/2, 0 ≤ C(t) ≤ 23.416 for (√2)/2 < t ≤ 2, 23.416 > C(t) ≥ 2
and for 0 ≤ t ≤ 0 1 ≤ B(t) ≤ 8 and for 1 < t ≤ 2, 8 > B(t) ≥ 1.5
Therefore, the curves intersect at in between (√2)/2 ≤ t ≤ 2.