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Câu hỏi: Tìm giá trị lớn nhất của hàm số $y=x-{{e}^{2x}}$ trên đoạn $\left[ -1;1 \right]$.
A. $\underset{\left[ -1;1 \right]}{\mathop{\max }} y=\dfrac{-\left( \ln 2+1 \right)}{2}$.
B. $\underset{\left[ -1;1 \right]}{\mathop{\max }} y=1-{{e}^{2}}$.
C. $\underset{\left[ -1;1 \right]}{\mathop{\max }} y=-\left( 1+{{e}^{-2}} \right)$.
D. $\underset{\left[ -1;1 \right]}{\mathop{\max }} y=\dfrac{\ln 2+1}{2}$.
A. $\underset{\left[ -1;1 \right]}{\mathop{\max }} y=\dfrac{-\left( \ln 2+1 \right)}{2}$.
B. $\underset{\left[ -1;1 \right]}{\mathop{\max }} y=1-{{e}^{2}}$.
C. $\underset{\left[ -1;1 \right]}{\mathop{\max }} y=-\left( 1+{{e}^{-2}} \right)$.
D. $\underset{\left[ -1;1 \right]}{\mathop{\max }} y=\dfrac{\ln 2+1}{2}$.
Ta có:
${f}'\left( x \right)=1-2{{e}^{2x}}=0\Rightarrow x=\dfrac{1}{2}\ln \dfrac{1}{2}\Rightarrow f\left( -1 \right)=1-\dfrac{1}{{{e}^{2}}};f\left( 1 \right)=1-{{e}^{2}};f\left( \dfrac{1}{2}\ln \dfrac{1}{2} \right)=-\dfrac{\ln 2+1}{2}$.
Vậy $\max y=-\dfrac{\ln 2+1}{2}$.
${f}'\left( x \right)=1-2{{e}^{2x}}=0\Rightarrow x=\dfrac{1}{2}\ln \dfrac{1}{2}\Rightarrow f\left( -1 \right)=1-\dfrac{1}{{{e}^{2}}};f\left( 1 \right)=1-{{e}^{2}};f\left( \dfrac{1}{2}\ln \dfrac{1}{2} \right)=-\dfrac{\ln 2+1}{2}$.
Vậy $\max y=-\dfrac{\ln 2+1}{2}$.
Đáp án A.