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alittlebitofcalculus-pdf-january2011-111112001007-phpapp01

Limit vt v10 vt v10 right derivative t 10 left

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Unformatted text preview: x) -- f (x) or δ x → 0 δx h We were actually taking the limit from the right. imit In other words, we were taking xL→ a+ i.e. approaching the instant a from the right. We should ask: what would happen if we took L imit i.e. approaching the L imit h→ 0 x → a-- instant a from the left ? f (x -- h ) -- f (x) or -- h because x → a-- i n terms of δ x is : this is L imit h→ 0 L imit δx → 0 f (x -- δ x) -- f (x) -- δ x x → a-- = a -- δ x as δ x → 0 and (a -- δ x ) -- a = -- δ x . Will we get the same INSTANTANEOUS RATE OF CHANGE ? In general, for any function f(x) we expect: INSTANTANEOUS RATE OF CHANGE INSTANTANEOUS RATE OF CHANGE = approaching a from the right approaching a from the left If they are equal then we say that f(x) is DIFFERENTIABLE at x = a. L imit x → a-- L imit δx → 0 f (x) -- f (a) = x -- a f (a -- δ x) -- f (a) = -- δ x 86 L imit x → a+ L imit δx → 0 f (x) -- f (a) x -- a f (a + δ x) -- f (a) δx Example 1: is f(x) differentiable at x = a ? 1 differentiable f(x) a f (x ) = { x for 0 ≤ x ≤ a for 2a -- x for a ≤ x ≤ 2 a a 0,0) (0,0 ) Approaching a from the left : f(x) = x left Near a and to the left of a : left 2a x x = a − δx f(x) − f(a) L imit f (a − δx) − f(a) = δx → 0 x−a (a − δx) − a L imit (a − δx) − a L imit − δx = δx → 0 = δx → 0 = +1 − δx − δx Limit x → a− Approaching a from the right : f(x) = 2a − x right Near a and to the right of a : x = a + δx right Limit x → a+ f(x) − f(a) L imit f (a + δx) − f(a) = δx → 0 x−a (a + δx) − a { 2a −(a +δx)} − { 2a −a } L L imit − δx = δximit0 = δx → 0 = −1 → + δx +δx f(x) − f(a) f(x) − f(a) left derivative Limit − ≠ right derivative Limit + right x→a x→a x−a x−a we say : f(x) is not differentiable at x = a. differentiable 87 Example 2 : is f(x) = |x| differentiable at x = 0 ? differentiable f(x) f(x) = |x| (0 , 0 ) Approaching 0 from the left : f(x) = |x| = −x left Near 0 and to the left...
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