Giải thích các bước giải:
$f(x)=\dfrac{\sqrt{4x^4+16x+17}}{2}\\ f'(x)=\dfrac{1}{2}.\dfrac{(4x^4+16x+17)'}{2\sqrt{4x^4+16x+17}}\\ =\dfrac{16x^3+16}{4\sqrt{4x^4+16x+17}}\\ =\dfrac{16(x^3+1)}{4\sqrt{4x^4+16x+17}}\\ f'(x)=0 \Leftrightarrow x^3+1=0 \Leftrightarrow x=-1\\ BBT:\\ \begin{array}{|c|ccccccccc|} \hline x&-\infty&&-1&&\infty\\\hline y'&&-&0&+&\\\hline &+\infty&&&&+\infty\\y&&\searrow&&\nearrow&\\&&&f(-1)\\\hline\end{array}\\ \Rightarrow min_{f(x)}=f(-1)$
