Từ đồ thị của \(y={f}'\left( x \right)\), ta có bảng biến thiên của \(y=f\left( x \right)\) trên \(\left[ -2;6 \right]\) như sau

Từ bảng biến thiên ta có

\(M=\underset{\left[ \text{-2;6} \right]}{\mathop{\text{max}}}\,\text{ }f\left( x \right)=\text{max}\left\{ f\left( 0 \right),f\left( 5 \right) \right\}, m=\underset{\left[ \text{-2;6} \right]}{\mathop{\text{min}}}\,\text{ }f\left( x \right)=\text{min}\left\{ f\left( -2 \right),f\left( 2 \right),f\left( 6 \right) \right\}\).

Từ đồ thị của \(y={f}'\left( x \right)\) ta có \(\int\limits_{0}^{2}{\left| {f}'\left( x \right) \right|\text{d}x}<\int\limits_{2}^{5}{\left| {f}'\left( x \right) \right|\text{d}x} \Leftrightarrow \int\limits_{0}^{2}{-{f}'\left( x \right)\text{d}x<\int\limits_{2}^{5}{{f}'\left( x \right)\text{d}x}} \Leftrightarrow f\left( 0 \right)-f\left( 2 \right)<0\)

Suy ra \(\text{max}\left\{ f\left( 0 \right),f\left( 5 \right) \right\}=f\left( 5 \right)\).

Mặt khác, cũng từ từ đồ thị của \(y={f}'\left( x \right)\), ta có \(\int\limits_{-2}^{0}{\left| {f}'\left( x \right) \right|\text{d}x}>\int\limits_{0}^{2}{\left| {f}'\left( x \right) \right|\text{d}x}\Leftrightarrow \int\limits_{-2}^{0}{{f}'\left( x \right)\text{d}x>\int\limits_{0}^{2}{-{f}'\left( x \right)\text{d}x}}\Leftrightarrow f\left( 0 \right)-f\left( -2 \right)>f\left( 0 \right)-f\left( 2 \right)\Leftrightarrow f\left( -2 \right)<f\left( 2 \right)\)

Hơn nữa \(\int\limits_{2}^{5}{\left| {f}'\left( x \right) \right|\text{d}x}>\int\limits_{5}^{6}{\left| {f}'\left( x \right) \right|\text{d}x} \Leftrightarrow \int\limits_{2}^{5}{{f}'\left( x \right)\text{d}x>\int\limits_{5}^{6}{-{f}'\left( x \right)\text{d}x}} \Leftrightarrow f\left( 5 \right)-f\left( 2 \right)>f\left( 5 \right)-f\left( 6 \right) \Leftrightarrow f\left( 2 \right)<f\left( 6 \right)\)

Suy ra \(\text{min}\left\{ f\left( -2 \right),f\left( 2 \right),f\left( 6 \right) \right\}=f\left( -2 \right)\).

Vậy \(\text{M=max}\left\{ f\left( 0 \right),f\left( 5 \right) \right\}=f\left( 5 \right), \text{m=min}\left\{ f\left( -2 \right),f\left( 2 \right),f\left( 6 \right) \right\}=f\left( -2 \right)\),

nên \(M+m=f\left( 5 \right)+f\left( -2 \right)\)