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):}`

Answer 1

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`