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Question
Verify the continuity and differentiability of f(x) = `{(1 - x if x < 1),((1 - x)(2 - x) if 1 <= x <= 2),(3 - x if x > 2):}` at x = 1 and x = 2.
Solution
`"L"[f(x)]_(x=1) = lim_(x->1) f(x) = lim_(h->0) "f"(1 - "h")`
`= lim_(h->0) 1 - (1 - "h")` [∵ f(x) = 1 - x if x < 1]
`= lim_(h->0)` h = 0 ...(1)
`"R"[f(x)]_(x=1) = lim_(x->1) f(x) = lim_(h->0) "f"(1 + "h")`
`= lim_(h->0)` [1 - (1 + h)][2 - (1 + h)] [∵ f(x) = (1 - x)(2 - x) if 1 ≤ x ≤ 2]`
`= lim_(h->0) (cancel(1) - cancel(1) - "h")(2 - 1 - "h")`
= `lim_(h->0)`(-h)(1 - h)
`= lim_(h->0)` - h + h2 = 0 + 0 = 0 ....(2)
Also f(1) = (1 - 1)(2 - 1) ...[∵ f(x) = (1 - x)(2 - x) when x = 1]
= 0 ...(3)
From (1), (2) and (3),
`"L"[f(x)]_(x=1) = "R"[f(x)]_(x=1)` = f(1) = 0
∴ f(x) is continuous at x = 1
`"L"["f"'(1)] = lim_(x->1^-) ("f"(x) - "f"(1))/(x - 1)`
`= lim_(h->0) ([1 - (1 - "h")] - [(1 - 1)(2 - 1)])/(1 - "h" - 1)`
`= lim_(h->0) ([1 - (1 - "h")] - [1 - (1 - "h")(2 - (1 - "h"))])/(-"h")` ...[∵ f(x) = (1 - x)(2 - x)]
`= lim_(h->0) (1 - 1 + "h")/(-"h")` = - 1 ....(4)
`"R"["f"'(1)] = lim_(x->1^+) ("f"(x) - "f"(1))/(x - 1)`
`= lim_(h->0) ("f"(1 + "h") - "f"(1))/(1 + "h" - 1)`
`= lim_(h->0) (1 - (1 + "h")(2 - (1 + "h")) - 0)/"h"`
`= lim_(h->0) ((-"h")(1 - "h"))/"h"` = - 1 ...(5)
From (4) and (5), `"L"["f"'(1)] = "R"["f"'(1)]`
∴ f(x) differentiable at x = 1
Now
`"L"[f(x)_(x=2)] = lim_(x->2^-) "f"(x) = lim_("h" -> 0)`f(2 - h)
`= lim_("h" -> 0) [1 - (2 - "h")][2 - (2 - "h")]` ...[f(x = (1 - x)(2 - x)) when x < 2]
`= lim_("h" -> 0)`(- 1 + h)(h)
`= lim_("h" -> 0)` = - h + h2 = 0 + 0 = 0 ....(6)
`"R"[f(x)_(x=2)] = lim_(x->2^+) "f"(x) = lim_("h" -> 0)`f(2 + h)
`= lim_(h->0)` 3 - (2 + h) ...[∵ f(x) = 3 - x when x > 2]
`= lim_(h->0)` 1 - h = 1 - 0 = 1
and f(2) = (1 - 2)(2 - 2) = 0 ....(7)
From (6) and (7), f(x) is not continuous at x = 2.
`"L"["f"'(2)] = lim_(x->2^-) ("f"(x) - "f"(2))/(x - 2)`
`= lim_(h->0) ("f"(2 - "h") - "f"(2))/(2 - "h" - 2)`
`= lim_(h->0) ([1 - (2 - "h")][2 - (2 - "h")] - [(1 - 2)(2 - 2)])/(-"h")` ....[∵ f(x = (1 - x)(2 - x)) when x < 2]
`= lim_(h->0) ((- 1 + "h")(+"h") - 0)/(- "h")`
`= lim_(h->0) (- 1 + "h")/(-1)`
= 1 ...(8)
`"R"["f"'(2)] = lim_(x->2^+) ("f"(x) - "f"(2))/(x - 2)`
`= lim_(h->0) ("f"(2 + "h") - "f"(2))/(cancel(2) + "h" - cancel(2))`
`= lim_(h->0) ([3 - (2 + "h")] - [(1 - 2)(2 - 2)])/"h"` ...[∵ f(x) = 3 - x when x > 2]
`= lim_(h->0) ((1 - "h") - 0)/"h"`
`= lim_(h->0) (1 - 0)/0 = ∞` ...(9)
`therefore "L"["f"'(2)] ne "R"["f"'(2)]`
⇒ f(x) is not differentiable at x = 2.
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