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Câu hỏi: Cho hàm số $y=f\left( x \right)$ có đạo hàm $f'\left( x \right)$ liên tục trên $\mathbb{R}$ và đồ thị hàm số $y=f'\left( x \right)$ trên đoạn $\left[ -2;6 \right]$ như hình vẽ.
Mệnh đề nào dưới đây đúng?
A. $\underset{\left[ -2;6 \right]}{\mathop{\max }} y=f\left( -1 \right)$.
B. $\underset{\left[ -2;6 \right]}{\mathop{\max }} y=f\left( 6 \right)$.
C. $\underset{\left[ -2;6 \right]}{\mathop{\max }} y=f\left( -2 \right)$.
D. $\underset{\left[ -2;6 \right]}{\mathop{\max }} y=f\left( 2 \right)$.
Dựa vào đồ thị trên ta có:
${{S}_{1}}>0\Leftrightarrow \int\limits_{-2}^{-1}{f'\left( x \right)}dx>0\Leftrightarrow \left. f\left( x \right) \right|_{-2}^{-1}>0\Leftrightarrow f\left( -1 \right)-f\left( -2 \right)>0\Leftrightarrow f\left( -1 \right)>f\left( -2 \right)$. (1)
${{S}_{1}}<{{S}_{2}}\Leftrightarrow \int\limits_{-2}^{-1}{f'\left( x \right)}dx<-\int\limits_{-1}^{2}{f'\left( x \right)}dx\Leftrightarrow \left. f\left( x \right) \right|_{-2}^{-1}<-\left. f\left( x \right) \right|_{-1}^{2}$
$\Leftrightarrow f\left( -1 \right)-f\left( -2 \right)<-\left( f\left( 2 \right)-f\left( -1 \right) \right)$
$\Leftrightarrow f\left( -1 \right)-f\left( -2 \right)<-f\left( 2 \right)+f\left( -1 \right)\Leftrightarrow f\left( -2 \right)>f\left( 2 \right)$. (2)
${{S}_{2}}<{{S}_{3}}\Leftrightarrow -\int\limits_{-1}^{2}{f'\left( x \right)}dx<\int\limits_{2}^{6}{f'\left( x \right)}dx\Leftrightarrow -\left. f\left( x \right) \right|_{-1}^{2}<\left. f\left( x \right) \right|_{2}^{6}$
$\Leftrightarrow -\left( f\left( 2 \right)-f\left( -1 \right) \right)<f\left( 6 \right)-f\left( 2 \right)$
$\Leftrightarrow -f\left( 2 \right)+f\left( -1 \right)<f\left( 6 \right)-f\left( 2 \right)\Leftrightarrow f\left( 6 \right)>f\left( -1 \right)$. (3)
Từ (1), (2) và (3) suy ra: $f\left( 6 \right)>f\left( -1 \right)>f\left( -2 \right)>f\left( 2 \right)$.
Vậy $\underset{\left[ -2;6 \right]}{\mathop{\max }} y=f\left( 6 \right)$.
A. $\underset{\left[ -2;6 \right]}{\mathop{\max }} y=f\left( -1 \right)$.
B. $\underset{\left[ -2;6 \right]}{\mathop{\max }} y=f\left( 6 \right)$.
C. $\underset{\left[ -2;6 \right]}{\mathop{\max }} y=f\left( -2 \right)$.
D. $\underset{\left[ -2;6 \right]}{\mathop{\max }} y=f\left( 2 \right)$.
${{S}_{1}}>0\Leftrightarrow \int\limits_{-2}^{-1}{f'\left( x \right)}dx>0\Leftrightarrow \left. f\left( x \right) \right|_{-2}^{-1}>0\Leftrightarrow f\left( -1 \right)-f\left( -2 \right)>0\Leftrightarrow f\left( -1 \right)>f\left( -2 \right)$. (1)
${{S}_{1}}<{{S}_{2}}\Leftrightarrow \int\limits_{-2}^{-1}{f'\left( x \right)}dx<-\int\limits_{-1}^{2}{f'\left( x \right)}dx\Leftrightarrow \left. f\left( x \right) \right|_{-2}^{-1}<-\left. f\left( x \right) \right|_{-1}^{2}$
$\Leftrightarrow f\left( -1 \right)-f\left( -2 \right)<-\left( f\left( 2 \right)-f\left( -1 \right) \right)$
$\Leftrightarrow f\left( -1 \right)-f\left( -2 \right)<-f\left( 2 \right)+f\left( -1 \right)\Leftrightarrow f\left( -2 \right)>f\left( 2 \right)$. (2)
${{S}_{2}}<{{S}_{3}}\Leftrightarrow -\int\limits_{-1}^{2}{f'\left( x \right)}dx<\int\limits_{2}^{6}{f'\left( x \right)}dx\Leftrightarrow -\left. f\left( x \right) \right|_{-1}^{2}<\left. f\left( x \right) \right|_{2}^{6}$
$\Leftrightarrow -\left( f\left( 2 \right)-f\left( -1 \right) \right)<f\left( 6 \right)-f\left( 2 \right)$
$\Leftrightarrow -f\left( 2 \right)+f\left( -1 \right)<f\left( 6 \right)-f\left( 2 \right)\Leftrightarrow f\left( 6 \right)>f\left( -1 \right)$. (3)
Từ (1), (2) và (3) suy ra: $f\left( 6 \right)>f\left( -1 \right)>f\left( -2 \right)>f\left( 2 \right)$.
Vậy $\underset{\left[ -2;6 \right]}{\mathop{\max }} y=f\left( 6 \right)$.
Đáp án B.