Let f:[-1,1]→R be defined as f(x) = ax² + bx + c for all x∈[-1,1], where a, b, c∈R such that f(-1)=2, f'(-1)=1 and for x∈(-1,1) the maximum value of f''(x) is 1/2. If f(x) ≤ α, for x∈[-1,1], then the least value of α is equal to ______.
Let f:[-1,1]→R be defined as f(x) = ax² + bx + c for all x∈[-1,1], where a, b, c∈R such that f(-1)=2, f'(-1)=1 and for x∈(-1,1) the maximum value of f''(x) is 1/2. If f(x) ≤ α, for x∈[-1,1], then the least value of α is equal to ______.
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1 Answer
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Given f (x) = ax^2 + bx + c.
f (-1) = a - b + c = 2
f' (x) = 2ax + b, so f' (-1) = -2a + b = 1
f' (x) = 2a, so f' (-1) = 2a = 1/2From 2a = 1/2, we get a = 1/4.
Substituting a into -2a + b = 1: -2 (1/4) + b = 1 => -1/2 + b = 1 => b = 3/2.
Substituting a and b into a - b + c = 2: 1/4 - 3/2 + c = 2 => -5/4 + c = 2 => c = 13/4.So, f (x) = (1/4)x^2 + (3/2)x + 13/4 = (1/4) (x^2 + 6x + 13).
We need to find f (1):
f (1) = (1/4) (1^2 + 6 (1) + 13) = (1/4) (1 + 6 + 13) = (1/4) (20) = 5.
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