a) \(\displaystyle y={{x}^{3}}-3{{x}^{2}}-9x+35\)
+) Xét \(\displaystyle D=\left[ -4;\ 4 \right]\) có :
\(\displaystyle y'=3{{x}^{2}}-6x-9\Rightarrow y'=0\Leftrightarrow 3{{x}^{2}}-6x-9=0\Leftrightarrow \left[ \begin{align} & x=3\ \in D \\ & x=-1\ \in D \\ \end{align} \right..\)
Ta có : \(\displaystyle y\left( -4 \right)=-41;\ \ y\left( 1 \right)=40;\ \ y\left( 3 \right)=8;\ \ y\left( 4 \right)=15.\)
Vậy \(\displaystyle \underset{x\in \left[ -4;\ 4 \right]}{\mathop{\max }}\,y=40\ \ khi\ \ x=-1\) và \(\displaystyle \underset{x\in \left[ -4;\ 4 \right]}{\mathop{\min }}\,y=-41\ \ khi\ \ x=-4.\)
+) Xét \(\displaystyle D=\left[ 0;\ 5 \right]\) có:
\(\displaystyle y'=3{{x}^{2}}-6x-9\Rightarrow y'=0\Leftrightarrow 3{{x}^{2}}-6x-9=0\Leftrightarrow \left[ \begin{align}& x=3\ \in D \\ & x=-1\ \notin D \\ \end{align} \right..\)
Ta có : \(\displaystyle y\left( 0 \right)=35;\ \ y\left( 3 \right)=8;\ \ y\left( 5 \right)=40.\)
Vậy \(\displaystyle \underset{x\in \left[ 0;\ 5 \right]}{\mathop{\max }}\,y=40\ \ khi\ \ x=5\) và \(\displaystyle \underset{x\in \left[ 0;\ 5 \right]}{\mathop{\min }}\,y=8\ \ khi\ \ x=3.\)
b) \(\displaystyle y={{x}^{4}}-3{{x}^{2}}+2\)
Ta có:\(\displaystyle y'=4{{x}^{3}}-6x\) \(\Rightarrow y'=0\Leftrightarrow 4{{x}^{3}}-6x=0\) \(\Leftrightarrow \left[ \begin{align}& x=0 \\ & x=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2} \\ & x=-\sqrt{\frac{3}{2}}=-\frac{\sqrt{6}}{2} \\ \end{align} \right.\)
+) Xét \(\displaystyle D=\left[ 0;\ 3 \right]\) có: \(\displaystyle x=-\frac{\sqrt{6}}{2}\notin D.\)
Có: \(\displaystyle y\left( 0 \right)=2;\ \ y\left( 3 \right)=56;\) \( y\left( \frac{\sqrt{6}}{2} \right)=-\frac{1}{4}.\)
Vậy \(\displaystyle \underset{x\in \left[ 0;\ 3 \right]}{\mathop{\min }}\,y=-\frac{1}{4}\ \ khi\ \ x=\frac{\sqrt{6}}{2}\) và \(\displaystyle \underset{x\in \left[ 0;\ 3 \right]}{\mathop{\max }}\,y=56\ \ khi\ \ x=3.\)
+) Xét \(\displaystyle D=\left[ 2;\ 5 \right]\) ta thấy \(\displaystyle x=0;\ \ x=\pm \frac{\sqrt{6}}{2}\ \ \notin \ D.\)
Có \(\displaystyle y\left( 2 \right)=6;\ \ y\left( 5 \right)=552.\)
Vậy \(\displaystyle \underset{x\in \left[ 2;\ 5 \right]}{\mathop{\min }}\,y=6\ \ khi\ \ x=2\) và \(\displaystyle \underset{x\in \left[ 2;\ 5 \right]}{\mathop{\max }}\,y=525\ \ khi\ \ x=5.\)
c) \(\displaystyle y=\frac{2-x}{1-x}=\frac{x-2}{x-1}\). Tập xác định: \(\displaystyle R\backslash \left\{ 1 \right\}.\)
Ta có: \(\displaystyle y'=\frac{1.\left( -1 \right)-1.\left( -2 \right)}{{{\left( x-1 \right)}^{2}}}=\frac{1}{{{\left( x-1 \right)}^{2}}}>0\ \ \forall x\ne 1.\)
+) Với \(\displaystyle D=\left[ 2;\ 4 \right]\) có: \(\displaystyle y\left( 2 \right)=0;\ \ y\left( 4 \right)=\frac{2}{3}.\)
Vậy \(\displaystyle \underset{x\in \left[ 2;\ 4 \right]}{\mathop{\min }}\,y=0\ \ khi\ \ x=2\) và \(\displaystyle \underset{x\in \left[ 2;\ 4 \right]}{\mathop{\max }}\,y=\frac{2}{3}\ \ khi\ \ x=4.\)
+) Với \(\displaystyle D=\left[ -3;\ -2 \right]\) có: \(\displaystyle y\left( -3 \right)=\frac{5}{4};\ \ y\left( -2 \right)=\frac{4}{3}.\)
Vậy \(\displaystyle \underset{x\in \left[ -3;\ -2 \right]}{\mathop{\min }}\,y=\frac{5}{4}\ \ khi\ \ x=-3\) và \(\displaystyle \underset{x\in \left[ -3;\ -2 \right]}{\mathop{\max }}\,y=\frac{4}{3}\ \ khi\ \ x=-2.\)
d) \(\displaystyle y=\sqrt{5-4x}\) . Tập xác định: \(\displaystyle \left( -\infty ;\ \frac{5}{4} \right].\)
Xét tập \(\displaystyle D=\left[ -1;\ 1 \right]:\)
Có: \(\displaystyle y'=\frac{\left( 5-4x \right)'}{2\sqrt{5-4x}}=\frac{-2}{\sqrt{5-4x}}<0\ \forall x\in \left[ -1;\ 1 \right].\)
Ta có: \(\displaystyle y\left( -1 \right)=3;\ \ y\left( 1 \right)=1.\)
Vậy \(\displaystyle \underset{x\in \left[ -1;\ 1 \right]}{\mathop{\min }}\,y=1\ \ khi\ \ x=1\) và \(\displaystyle \underset{x\in \left[ -1;\ 1 \right]}{\mathop{\max }}\,y=3\ \ khi\ \ x=-1.\)
