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Question
Determine the values of p and q that make the function f(x) differentiable on R where
f(x) `{:( = "p"x^3",", "for" x < 2),(= x^2 + "q"",", "for" x ≥ 2):}`
Solution
f(x) `{:( = "p"x^3",", x < 2),(= x^2 + "q"",", x ≥ 2):}`
Continuity at x = 2:
f(x) is continuous at x = 2
∴ `lim_(x -> 2^-) "f"(x) = lim_(x -> 2^+) "f"(x)`
∴ `lim_(x -> 2^-) "p"x^3 = lim_(x -> 2^+) (x^2 + "q")`
∴ 8p = 4 + q
∴ 8p – q = 4 ...(i)
Differentiability at x = 2:
`lim_("h" -> 0^-) ("p"(2 + "h")^3 - (4 + "q"))/"h"`
= `lim_("h" ->0^+) ((2 + "h")^2 + "q" - (4 + "q"))/"h"`
∴ `lim_("h" -> 0^-) [("ph"^3 + 6"ph"^2 + 12"hp" + 8"p" - (4 + "q"))/"h"]`
= `lim_("h" -> 0) [("h"^2 + 4"h")/"h"]`
∴ `lim_("h" -> 0^-) ("ph"^2 + 6"ph" + 12"p") = lim_("h" -> 0) ("h" + 4)`
∴ 12p = 4
∴ p = `1/3`
Substituting p = `1/3` in (i), we get
q = `8/3 - 4 = -4/3`
∴ p = `1/3`, q = `-4/3`
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