Răspuns:
f(x)≥14
Explicație pas cu pas:
f:R->R, f(x)=√(4x²+28x+85) + √(4x²-28x+113).
4x²+28x+85=(4x²+28x+49)+36=(2x+7)²+6²
4x²-28x+113= (4x²-28x+49)+64=(2x-7)²+8²
f(x)=√ [(2x+7)²+6²] +√[(2x-7)²+8²]
(2x+7)²≥0, (2x-7)²≥0
√ [(2x+7)²+6²]≥√6²=6
√ [(2x-7)²+8²]≥√8²=8
=> f(x)≥6+8
f(x)≥14
