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$1/$ Hàm số xác định trên đoạn $\left[ {\frac{1}{2},\,4} \right]$
Ta có $f(x) = \left\{ \begin{array}{l}
{y_1} = - {x^2} - 2x + 3 + \frac{3}{2}\ln x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{2} \le x \le 1\\
{y_2} = {x^2} + 2x - 3 + \frac{3}{2}\ln x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1 \le x \le 4
\end{array} \right.$
* Với $\frac{1}{2} \le x \le 1:\,\,\,\,\,\,\,{f' }\left( x \right) = {y_1}' = - 2x - 2 + \frac{3}{{2x}}$
${f' }\left( x \right) = \frac{{\left( { - 2x + 1} \right)\left( {2x + 3} \right)}}{{2x}} < 0\,\,\,\,\,\,\,\,\,\,\forall x \in \left[ {\frac{1}{2},1} \right]$
* Với $1 \le x \le 4:\,\,\,\,\,\,{f'}\left( x \right) = {y_2}' = 2x + 2 + \frac{3}{{2x}} > 0\,\,\,\,\forall x \in \left[ {1,4} \right]$
Ta có $f\left( {\frac{1}{2}} \right) = \frac{7}{4} - \frac{3}{2}\ln 2$
$\begin{array}{l}
f(1)\,\,\,\,\, = 0\\
f(4)\,\,\,\, = 21 + 3\ln 2
\end{array}$
Suy ra : $\mathop {Maxf(x)}\limits_{\frac{1}{2} \le x \le 4} =f(4)= 21 + 3\ln 2$
$\mathop {\min f(x)}\limits_{\frac{1}{2} \le x \le 4} =f(1)= 0$
$2)$ Hàm số xác định trên đoạn $[\frac12;2]$
Ta có:
$f(x)=\left\{ \begin{array}{l} -x^2-x+2+\ln x khi \frac 1 2 \leq x\le 1 \\ x^2+x-2+\ln x khi 1\le x\le 2 \end{array} \right.$
Với $\frac 1 2\le x\le 1$ ta có: $f'(x)=-2x-1+\frac 1 x\le 0 \forall x\in[\frac 1 2;1]$
Với $1\le x\le 2$ ta có: $f'(x)=2x+1+\frac 1 x>0 \forall x\in[1;2]$
Ta có: $f(\frac 1 2)=\frac 5 4 +\ln (\frac 1 2); f(1)=0; f(2)=4+\ln 2$
Vậy: $\mathop {Maxf(x)}\limits_{\frac{1}{2} \le x \le 2} = f(2)=4 + \ln 2$
$\mathop {minf(x)}\limits_{\frac{1}{2} \le x \le 2} =f(1)= 0$
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