Mighty Min

Algebra Level 3

Consider the function defined as

f ( x ) = ( x 2 + 2 x + 3 ) 2 + 4. f(x) = (x^2 + 2x + 3)^2 + 4.

As x x ranges over all real values, what is the minimum value of f ( x ) f(x) ?


The answer is 8.

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1 solution

Arron Kau Staff
May 13, 2014

We know that squares are always non-negative. This implies that f ( x ) 4 f(x) \geq 4 directly. However, this does not imply that the minimum value must be 4.

In fact, we know that x 2 + 2 x + 3 = ( x + 1 ) 2 + 2 x^2 + 2x + 3 = (x+1)^2 + 2 . Hence, ( x 2 + 2 x + 3 ) 2 (x^2 + 2x + 3)^2 achieves a minimum value of 2 2 2^2 when x = 1 x = -1 . As such, f ( x ) 2 2 + 4 = 8 f(x) \geq 2^2 + 4 = 8 , with equality when x = 1 x = -1 .

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