Let the function, ƒ:[-7, 0]$\to$R be continuous on [-7,0] and differentiable on (-7, 0). If ƒ(-7) = - 3 and ƒ'(x) $\le$ 2, for all x $\in$ (-7,0), then for all such functions ƒ, ƒ(-1) + ƒ(0) lies in the interval:

Using Lagrange’s Mean Value Theorem in [–7, –1] <br><br>$${{f\left( { - 1} \right) - f\left( { - 7} \right)} \over { - 1 - \left( { - 7} \right)}}$$ = f'(c₁) <br><br>As ƒ'(x) $\le$ 2 then f'(c₁) $\le$ 2 <br><br>$\therefore$ $${{f\left( { - 1} \right) - f\left( { - 7} \right)} \over { - 1 - \left( { - 7} \right)}}$$ $\le$ 2 <br><br>$\Rightarrow$ ${{f\left( { - 1} \right) + 3} \over 6}$ $\le$ 2 <br><br>$\Rightarrow$ f(-1) $\le$ 9 <br><br>Using Lagrange’s Mean Value Theorem in [–7, 0] <br><br>$\Rightarrow$ ${{f\left( 0 \right) - f\left( { - 7} \right)} \over {0 - \left( { - 7} \right)}}$ $\le$ 2 <br><br>$\Rightarrow$ f(0) $\le$ 11 <br><br>$\therefore$ ƒ(-1) + ƒ(0) $\le$ 11 + 9 <br><br>$\Rightarrow$ ƒ(-1) + ƒ(0) $\le$ 20