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If f(x) = 16x^2 + 19x - 32 what is the minimum value of f(x).
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The function f(x) = 16x^2 + 19x - 32. At the point where the value of f(x) is the least f'(x) = 0. In addition, for the solution x = a of f'(x) = 0, the value of f''(a) should be positive.
f(x) = 16x^2 + 19x - 32
f'(x) = 32x + 19
f''(x) = 32
f'(x) = 0
=> x = -19/32
The value of f''(x) is positive for all values of x.
At x = -19/32, f(x) = -2406/64
The minimum value of f(x) is -2406/64
Posted by justaguide on July 15, 2013 at 3:14 PM (Answer #1)
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