The absolute minimum = -2√2.
The absolute maximum= 4.5
Consider f(t)=t√9-t on the interval (1,3].
Find the critical points: Find f'(t)=0.
f"(t) = 0
√9-t² d/dt t + t d/dt √9-t²=0
√9-t² + t/2√9-t² (-2t)=0
9-t²-t²/√9-t²=0
9-2t²=0
9=2t², t²=9/2, t=±3/√2
since -3/√2∉ (1,3].
Therefore, the critical point in the interval (1,3] is t= 3/√2.
Find the value of the function at t=1, 3/√2,3 to find the absolute maximum and minimum.
f(-1)=-1√9-1²
= -√8 , =-2√2
f(3/√2)= 3/√2 √9-(3/√2)²
= 3/√2 √9-9/2
=3/√2 √9/2
=9/2 = 4.5
f(3)= 3√9-3²
= 3(0)
=0
The absolute maximum is 4.5 and the absolute minimum is -2√2.
The absolute maximum point is the point at which the function reaches the maximum possible value. Similarly, the absolute minimum point is the point at which the function takes the smallest possible value.
A relative maximum or minimum occurs at an inflection point on the curve. The absolute minimum and maximum values are the corresponding values over the full range of the function. That is, the absolute minimum and maximum values are bounded by the function's domain.
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Answer:
The total number of samples that give this outcome is 5.
Step-by-step explanation:
Since Y takes values in {0,1,2,3}, For us to have that 
implies that all of them are zero but one. The one that is non-zero necessarily is equal to 1. To calculate the number of samples that give this outcome is equivalent to counting the total number of ways in which we can pick the i-index such that 
. Note that in this case we can either choose Y1 to be 1, Y2 to be 1 and so on. So, the total number of samples that give this outcome is 5.
Answer:
45 meters
Step-by-step explanation:
If x represents the seconds after the launch, then the time of launch is when x=0 so you just need to solve for h(0)
h(0) = -5(1)(-9)
h(0) = 45
Answer:
I think the answer is 3
Step-by-step explanation:
