If $A_G$ is the adjacency matrix of a k-regular graph, let $B = J+xA_G$, where J is the matrix whose elements are all 1s and $x\in R$ is a scalar. If $\lambda_1\geq\lambda_2\geq \dots \geq \lambda_n$ are eigenvalues of $A_G$, how do we prove that $\underset{x\in R}{\min} \lambda_{max}(J+xA_G) = \frac{n\lambda_n}{\lambda_n-\lambda_1}$?

I know that e (the vector of all 1s) is an eigenvector of B with eigenvalue $n+x\lambda_1$, but what about the other eigenvectors? If we know other eigenvectors of B, then we can compare among them and pick the maximum. Thanks!