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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Step-by-step explanation:
it's this > with _ under it
Answer:
Step-by-step explanation:
so you dont know what m is so lets leave that to the side.
first you see its -4 and six. so m = 4 and thats the naswer
I turn the mixed numbers into decimals before doing anything else.
11 2/5----> 57/5----> 11.4
4 2/7----> 30/7----> 4.29
2 1/7----> 15/7----> 2.14
11.4=q-4.29+2.14
11.4=q-2.15
13.55= q
13.55 converted to a fraction is 13 11⁄20
q=13 11⁄20
Step-by-step explanation:
the area = 
= 
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