Suppose that $ f:\mathbb{R}^n \to \mathbb{R}$ is given by $f(x)=a_1x_1^2+a_2x_2^2+..........+a_nx_n^2,$ where $ x=(x_!,x_2,....,x_n)$ and at least one $a_j$ is not zero. Then we can conclude that
f is not everywhere differentiable
the gradient $(\nabla f)(x)\ne 0$ for every $ x \in \mathbb R^n$
if $ x \in \mathbb R^n$ is such that $(\nabla f)(x)=0 $ then $f(x)=0$
if $ x \in \mathbb R^n$ is such that $f(x)=0$ then $(\nabla f)(x)=0 $
My attempt: $1.\;\; f'(x)=2a_1x_1+2a_2x_2+.......+2a_nx_n \Rightarrow$ f is differentiable. hence 1. is false
Let $f(x)=x_1^2+x_2^2$ then $(\nabla f)(x)=2x_1+2x_2$. when $x=(1,-1)$, $(\nabla f)(x)=0$. hence 2. is false
from (2) we have $(\nabla f)(x)=0$ but $f(x)=(1)^2+(-1)^2 \ne 0$. hence $(\nabla f)(x)=0$ does not imply that $f(x)=0$
Let $f(x)=x_1^2-x_2^2$ then $(\nabla f)(x)=2x_1-2x_2$. when $x=(1,-1), f(x)=0 $but $(\nabla f)(x) \ne 0$. hence $f(x)=0$ does not imply that $(\nabla f)(x)=0$
Is my explaination correct? In the key answer it is given that option 3 is true. I am not understanding how the option 3 is correct. please help me to understand this concept