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Question
For what value of λ is the function defined by f(x) = `{(λ(x^2 - 2x)", if" x <= 0),(4x+ 1", if" x > 0):}` continuous at x = 0? What about continuity at x = 1?
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Solution
f(x) = `{(λ(x^2 - 2x)", if" x <= 0),(4x+ 1", if" x > 0):}`
If f(x) is continuous at x = 0, it implies:
f(0) = `lim_(x -> 0^+)` f(x) = `lim_(x -> 0^-)` f(x)
⇒ [02 − 2(0)] = [4(0) + 1] = [02 − 2(0)]
⇒ 0 = 1 = 0
Which cannot be true, i.e. for any value of λ this function is not continuous at x = 0.
If f(x) is continuous at x = 1, this implies:
f(1) = `lim_(x -> 1^+)` f (x) = `lim_(x -> 1^-)` f(x)
⇒ 4(1) + 1 = 4(1) + 1 = 4(1) + 1
⇒ 5 = 5 = 5
Which is always true, i.e. for any value of λ this function is continuous at x = 1.
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