Question

Find the absolute minimum and absolute maximum values of f on the given interval. f(t) = t 9 − t2 , [−1, 3]

1. 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.