Đáp án:
$1/ \text{Min} = - 3$
$2/ a,Minf(x)_{[\frac{1}{2};2]} = 3 \Leftrightarrow x =1$
$b, Minf(x)_{[\frac{1}{5};\frac{4}{5}]} = 3 \Leftrightarrow x =1$
Giải thích các bước giải:
1/ $\sqrt{3x^2+1}+\sqrt{9-x^2}$
ĐKXĐ : $\begin{cases} 3x^2+1\ge0\\9-x^2\ge0\end{cases}$
$\begin{cases} x^2\ge\dfrac{-1}{3}(\text{luôn đúng}) \\|x|\le3\end{cases}$
$\to \left[ \begin{array}{l}x\le3\\x\ge-3 \end{array} \right.$
Vậy $\text{min} = - 3$
2/ $f'(x) = 2x-\dfrac{2}{x^2} =0$
$\Leftrightarrow x=1$
+ $x \in [\frac{1}{2};2]$
$f(1) = 3$
$f(\dfrac{1}{2})= \dfrac{17}{4}$
$f(2)=5$
$\to Minf(x)_{[\frac{1}{2};2]} = 3 \Leftrightarrow x =1$
+ $x \in [\dfrac{1}{5};\dfrac{4}{5}]$
$f(1)= 3$
$f(\dfrac{1}{5})=\dfrac{251}{25}$
$f(\dfrac{4}{5})=\dfrac{157}{50}$
$\to Minf(x)_{[\frac{1}{5};\frac{4}{5}]} = 3 \Leftrightarrow x =1$
