0 0 votes Let $f(.)$ be a twice differentiable function from $\mathbb{R}^2 \rightarrow \mathbb{R}$. If $\boldsymbol{p}, x_0 \in \mathbb{R}^2$ where $\|p\|$ is sufficiently small (here $|| . ||$ is the Euclidean norm or distance function), then $f\left(\boldsymbol{x}_{\mathbf{0}}+\boldsymbol{p}\right)=f\left(\boldsymbol{x}_{\mathbf{0}}\right)+\nabla f\left(\mathbf{x}_{\mathbf{0}}\right)^{\mathrm{T}} \boldsymbol{p}+\frac{1}{2} \boldsymbol{p}^T \nabla^2 f(\boldsymbol{\psi}) \boldsymbol{p}$ where $\boldsymbol{\psi} \in \mathbb{R}^2$ is a point on the line segment joining $x_0$ and $x_0+\boldsymbol{p}$. If $x_0$ is a strict local minimum of $f(\boldsymbol{x})$, then which one of the following statements is TRUE? $\nabla f\left(\mathbf{x}_{0}\right)^{\mathrm{T}} \boldsymbol{p}>0$ and $\boldsymbol{p}^{T} \nabla^{2} f(\boldsymbol{\psi}) \boldsymbol{p}=0$ $\nabla f\left(\mathbf{x}_{0}\right)^{\mathrm{T}} \boldsymbol{p}=0$ and $\boldsymbol{p}^{T} \nabla^{2} f(\boldsymbol{\psi}) \boldsymbol{p}>0$ $\nabla f\left(\mathbf{x}_{0}\right)^{\mathrm{T}} \boldsymbol{p}=0$ and $\boldsymbol{p}^{T} \nabla^{2} f(\boldsymbol{\psi}) \boldsymbol{p}=0$ $\nabla f\left(\mathbf{x}_{0}\right)^{\mathrm{T}} \boldsymbol{p}=0$ and $\boldsymbol{p}^{T} \nabla^{2} f(\boldsymbol{\psi}) \boldsymbol{p}<0$ Others gateme-2024 calculus engineering-mathematics + – ♦admin 590 views answer comment Share Follow Add Sync Questions 0 reply Please log in or register to add a comment.