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
[tex]C(t) = t\times e^{4-t^2} \ for \ 0\leq t\leq 2[/tex]
Finding the boundaries of the function gives;
[tex]0\times e^{4-0^2} = 0[/tex] and [tex]2\times e^{4-2^2} = 2[/tex]
At t = 1, we have [tex]1\times e^{4-1^2} = e^{3} = 20.086[/tex]
We find the maximum point as follows;
[tex]\frac{\mathrm{d} \left (t\times e^{4-t^2} \right )}{\mathrm{d} x}=0[/tex]
From which we have;
[tex]\frac{\mathrm{} e^{4-t^2} - t\times e^{4-t^2} \times2\times t }{(e^{4-t^2} )^2}=0[/tex]
[tex]e^{4-t^2} - t\times e^{4-t^2} \times2\times t }=0[/tex]
[tex]e^{4-t^2}(1 - t\times2\times t })=0\\e^{4-t^2}(1 - 2\times t^2 })=0\\[/tex]
[tex]e^{4-t^2}=0[/tex] or [tex](1 - 2\times t^2 })=0[/tex]
∴ 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.
