Problem: For x≧0 the smallest value of 6(1+x)4x2+8x+13 is:
Answer Choices:
A. 1
B. 2
C. 1225
D. 613
E. 534
Solution:
Let y=6(1+x)4x2+8x+13∴4x2+(8−6y)x+13−6y=0, a quadratic equation in x with discriminant
D=36(y2−4). Since x≧0,D must be non-negative.
∴y≧2, that is, the minimum value of the given fraction is 2 .
Let y=6(1+x)4x2+8x+13=32(x+1)+2(x+1)3 so that y is of the form N+N1. The minimum value of such an expression occurs when N=N1, that is, when 32(x+1)=2(x+1)3, in other words, when x=21. With this value of x,y=2.