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.
Learn more about Absolute minimum and maximum here:brainly.com/question/19921479
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The first is C
The second is D
The third is A
Answer:
(-14+3/2b) - (1+8/2b) goes with -15-5/2b
(5+26)+2b+3/2) goes with 4b+13/2
(-10+b)+(7b-5) goes with 8b-15
(7/2b-3)-(8+6b) goes with -5/2b-11
Step-by-step explanation:
Answer: 12 only
Step-by-step explanation:
The given graph is representing the function f(x).
To find the value of f(0), we need to check the position of curve at x=0.
It implies we need to check the position of curve on y-axis.
We can see there is only one point where function curve is intersection the y-axis such that at x=0 .i.e. y=12
Hence, from the given graph the value of 
only.
