Question

For what value of λ is the function 
\[f\left( x \right) = \begin{cases}\lambda( x^2 - 2x), & \text{ if }  x \leq 0 \\ 4x + 1 , & \text{  if } x > 0\end{cases}\]continuous at x = 0? What about continuity at x = ± 1?

Answer 1

The given function f is \[f\left( x \right) = \begin{cases}\lambda( x^2 - 2x), & \text{ if }  x \leq 0 \\ 4x + 1 , & \text{  if } x > 0\end{cases}\]

If f is continuous at x = 0, then

Therefore, there is no value of λ for which f(x) is continuous at x = 0.

At x = 1,

f (1) = 4x + 1 = 4 × 1 + 1 = 5

Therefore, for any values of λ, f is continuous at x = 1

At x = -1, we have

f (-1) =

\[\lambda\left( 1 + 2 \right) = 3\lambda\]

\[\lim_{x \to - 1} \lambda\left( 1 + 2 \right) = 3\lambda\]

\[ \therefore \lim_{x \to - 1} f\left( x \right) = f\left( - 1 \right)\]

Therefore, for any values of λ,  f is continuous at x = -1