For what value of λ is the function defined by f(x) = \(\begin{cases} \lambda(x^2 - 2x) & if\; x \;\leq0\\ 4x + 1, & if\; x > 0 \end{cases}\)

Question

For what value of λ is the function defined by \(\begin{cases} \lambda(x^2 - 2x) & if\; x \;\leq0\\ 4x + 1, & if\; x > 0 \end{cases}\) continous at x = 0? 

what about continuity at x = 1?

Answer 1

∴ For any value of λ, the function is discontinuous, how ever at x = 1 and x = -1. 

we can prove that the function is continuous.