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Câu hỏi: Tìm giá trị lớn nhất của hàm số $f\left( x \right)=\dfrac{x}{x+2}$ trên đoạn $\left[ 1;4 \right]$.
A. $\underset{\left[ 1;4 \right]}{\mathop{max}} f\left( x \right)=4$.
B. $\underset{\left[ 1;4 \right]}{\mathop{max}} f\left( x \right)=1$.
C. $\underset{\left[ 1;4 \right]}{\mathop{max}} f\left( x \right)=\dfrac{1}{3}$.
D. $\underset{\left[ 1;4 \right]}{\mathop{max}} f\left( x \right)=\dfrac{2}{3}$.
A. $\underset{\left[ 1;4 \right]}{\mathop{max}} f\left( x \right)=4$.
B. $\underset{\left[ 1;4 \right]}{\mathop{max}} f\left( x \right)=1$.
C. $\underset{\left[ 1;4 \right]}{\mathop{max}} f\left( x \right)=\dfrac{1}{3}$.
D. $\underset{\left[ 1;4 \right]}{\mathop{max}} f\left( x \right)=\dfrac{2}{3}$.
Ta có: ${f}'\left( x \right)=\dfrac{2}{{{\left( x+2 \right)}^{2}}}>0,\forall x\in \left[ 1;4 \right]$
$\underset{\left[ 1;4 \right]}{\mathop{\Rightarrow max}} f\left( x \right)=f\left( 4 \right)=\dfrac{4}{6}=\dfrac{2}{3}$
$\underset{\left[ 1;4 \right]}{\mathop{\Rightarrow max}} f\left( x \right)=f\left( 4 \right)=\dfrac{4}{6}=\dfrac{2}{3}$
Đáp án D.