One Step on the Input, Not the Weights

A two-layer network with a ReLU hidden layer computes

$\mathbf{z} = W_1\mathbf{x} + \mathbf{b}_1, \qquad \mathbf{h} = \mathrm{ReLU}(\mathbf{z}), \qquad \hat{\mathbf{y}} = W_2\mathbf{h} + \mathbf{b}_2, \qquad L = \tfrac{1}{2}\left\lVert \hat{\mathbf{y}} - \mathbf{t} \right\rVert^2$

with

$W_1 = \begin{bmatrix} 1 & 2 \\ -2 & 1 \\ 1 & -1 \end{bmatrix}, \quad \mathbf{b}_1 = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}, \qquad W_2 = \begin{bmatrix} 1 & -3 & 2 \\ 2 & -1 & 1 \end{bmatrix}, \quad \mathbf{b}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$

The input and target are

$\mathbf{x} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \qquad \mathbf{t} = \begin{bmatrix} 5 \\ 7 \end{bmatrix}$

$\mathrm{ReLU}$ acts entrywise, $\mathrm{ReLU}(u) = \max(0,\ u)$, with derivative $1$ where $u > 0$ and $0$ where $u < 0$. No pre-activation in this problem is ever exactly zero.

The weights are frozen. Instead of training the network, one gradient-descent step is taken on the input itself:

$\mathbf{x}' = \mathbf{x} - \eta\,\frac{\partial L}{\partial \mathbf{x}}, \qquad \eta = 0.02$

The updated input $\mathbf{x}'$ is then fed through the same unchanged network.

What is the new loss $L(\mathbf{x}')$? Round your answer to 3 decimal places.