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For what value of lambda is the functio...

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`?

Text Solution

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`f(x)` to be continuous at `x = 0`,
`f(0^-) = f(0^+)`
`=> lambda(0-0) = 4(0)+1`
`0 = 1,` which is not true.
So, `f(x)` is not continuots at `x = 0`.
Now, at `x = 1`,
`Lim_(x->1) f(x) = Lim_(x->1) 4x+1 = 5 = f(1)`
Therefore, `f(x)` is continuous at `x = 1`.
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