The maximum value of quadratic function in closed interval Y = T ^ 2-2t + 3 - 1 / 4 ≤ t ≤ 2 find the maximum and minimum of Y | Math enthusiast | Mathematical art

The maximum value of quadratic function in closed interval Y = T ^ 2-2t + 3 - 1 / 4 ≤ t ≤ 2 find the maximum and minimum of Y

y=t²-2t+3=(t-1)²+2 -1/4≤t≤2
The axis of symmetry is t = 1 with the opening upward
So the function decreases monotonically in [- 1 / 4,1] and increases monotonically in [1,2]
therefore
When t = 1, the minimum value of the function is 2
When t = - 1 / 4, the function gets the maximum value (- 1 / 4-1) &# 178; + 2 = 25 / 16 + 2 = 57 / 16

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