# 4thsol - Calculus 119 4th Week 1 1 f x = x 2 1 if 1< x<...

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Unformatted text preview: Calculus 119 4th Week 1 1. f ( x ) = (- x 2 + 1 if- 1 < x < 1 x 2- 1 otherwise f ( x ) is differentiable everywhere except the points -1 and 1, and the derivative of f ( x ) is given below; f ( x ) =- 2 x when- 1 < x < 1 f ( x ) = 2 x when x <- 1 and x > 1 Does f ( x ) exist at the points -1 and 1? If it exists, what? At the point -1 lim x →- 1 + f ( x )- f (- 1) x + 1 = lim x →- 1 +- x 2 + 1- x + 1 = lim x →- 1 + 1- x = 2 lim x →- 1- f ( x )- f (- 1) x + 1 = lim x →- 1- x 2- 1- x + 1 = lim x →- 1- x- 1 =- 2          ⇒ lim x →- 1 f ( x )- f (- 1) x +1 does not exist. At the point 1 lim x → 1 + f ( x )- f (1) x- 1 = lim x → 1 + x 2- 1- x- 1 = lim x → 1 + x + 1 = 2 lim x → 1- f ( x )- f (1) x- 1 = lim x → 1-- x 2 + 1- x- 1 = lim x → 1-- (1 + x ) =- 2          ⇒ lim x → 1 f ( x )- f (1) x- 1 does not exist. Since f ( x ) does not exist at the points -1 and 1, The point -1 and 1 are critical points (or numbers). Is there other critical point? When x <- 1 and x > 1, f ( x ) = 2 x = 0 ⇒ x = 0. Since x / ∈ (-∞ ,- 1) ∪ (1 , ∞ ), x = 0 is not critical point. On the other hand, 1 Please email all corrections and suggestions to these solutions to [email protected] email all corrections and suggestions to these solutions to [email protected]
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4thsol - Calculus 119 4th Week 1 1 f x = x 2 1 if 1< x<...

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