For what value of `lambda`is the function defined by `f(x)={lambda(x^2-2x), ifxlt=0 4x+1, ifx >0`continuous at `x = 0`? What about continuity at `x = 1`?

Explanation

`f(x)={{:(lambda(x^(2)-2x )"," if x le 0),(4x + 1 "," if x gt 0):}`
at x = 0
f(0) `lambda` (0 - 0) = 0
R.H.L. = ` underset (x to 3^(+))(lim)f(x)`
Let x = 0 + h
`rArr 0 + h to 0`
`rArr h to 0`
`=underset (h to 0)(lim) f(0+h)`
`=underset (h to 0)(lim) 4(0+h)+1`
`= 4 (0 + 0) + 1 = 1 `
`:'` f(0) is not continuous at x = 0 for any value of `lambda`
For x `gt` 0, f(x) = 4x + 1 is a polynomial function which is continuous.
`:.` f(x) is continuous at x = 1 for all value of `lambda`