<s>[INST] If $x\geq 0,$ what is the smallest value of the function $f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$ If $x\geq 0,$ what is the smallest value of the function
$$f(x)= \frac{4x^2+ 8x + 13}{6(1+ x)}$$
I tried doing it by completing the square in numerator and making it of the form
$$\frac{4(x+ 1)^2+ 9}{6(1+ x)}$$
and then, I put the value of $x= 0$ and the answer is coming out to be $13/6.$
But the actual answer is $2.$
Am I missing something ?
 [/INST] The derivative, $$\frac{2}{3}-\frac{3}{2 (x+1)^2}$$ is zero when $x=1/2$:

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