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Find the value of k so that the function f is continuous at the indicated point.
f(x) = `{(kx + 1", if" x <= 5),(3x - 5", if" x > 5):}` at x = 5
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f(x) = `{(kx + 1", if" x <= 5),(3x - 5", if" x > 5):}`
If f(x) is continuous at x = 5, this implies:
f(5) = `lim_(x -> 5^+)` f(x) = `lim_(x -> 5^-)` f(x)
⇒ k(5) + 1 = 3(5) − 5
⇒ 5k + 1 = 15 − 5
⇒ 5k + 1 = 10
⇒ 5k = 10 − 1
⇒ 5k = 9
⇒ k = `9/5`
That is, for the quantity k = `9/5` this function is continuous at x = 5.
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