## Thomas' Calculus 13th Edition

$u_{max}= \lt -\dfrac{1}{\sqrt 2}, \dfrac{1}{\sqrt 2} \gt$ and $u_{min}= \lt \dfrac{1}{\sqrt 2}, -\dfrac{1}{\sqrt 2} \gt$ $D_u f_{max} =\sqrt 2 \\ D_u f_{min} =- \sqrt 2$
Since, $D_u f = \nabla f \cdot u$ $$f_x (-1,1) = 2x+y=(2)(-1)+1=-1 \\ f_y (-1,1) =x+2y=-1+(2)(1)=1$$ and $u_{max}=\dfrac{\nabla f (-1,1)}{|\nabla f (-1,1) |}= \lt -\dfrac{1}{\sqrt 2}, \dfrac{1}{\sqrt 2} \gt$ $\implies u_{min}=- u_{max}= \lt \dfrac{1}{\sqrt 2}, -\dfrac{1}{\sqrt 2} \gt$ Now, $D_u \ f_{max} =\nabla f (-1,1) \cdot u_{max} \\=\lt -1, 1 \gt \cdot \lt -\dfrac{1}{\sqrt 2}, \dfrac{1}{\sqrt 2} \gt \\ =\sqrt 2$ and $D_u \ f_{min} =\nabla f (-1,1) \cdot u_{min} \\=\lt -1, 1 \gt \cdot \lt \dfrac{1}{\sqrt 2}, -\dfrac{1}{\sqrt 2} \gt \\ =- \sqrt 2$