For what value of $\lambda$ is the function defined by $f(x) = \begin{cases} \lambda(x^2 - 2x), & \text{if } x \le 0 \\ 4x + 1, & \text{if } x > 0 \end{cases}$ continuous at $x=0$? What about continuity at $x=1$?

(NONE) For the function $f(x)$ to be continuous at $x=0$,the left-hand limit,right-hand limit,and the value of the function at $x=0$ must be equal.
Left-hand limit at $x=0$:
$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \lambda(x^2 - 2x) = \lambda(0^2 - 2(0)) = 0$
Right-hand limit at $x=0$:
$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (4x + 1) = 4(0) + 1 = 1$
Value of the function at $x=0$:
$f(0) = \lambda(0^2 - 2(0)) = 0$
Since the left-hand limit $(0)$ is not equal to the right-hand limit $(1)$,the function is discontinuous at $x=0$ for any value of $\lambda$.
At $x=1$,the function is defined by $f(x) = 4x + 1$,which is a polynomial function. Polynomial functions are continuous everywhere in their domain. Thus,$f(x)$ is continuous at $x=1$ for any value of $\lambda$.