Question

f(x) = `{{:(x^2/2",",  "if"  0 ≤ x ≤ 1),(2x^2 - 3x + 3/2",",  "if"  1 < x ≤ 2):}` at x = 1

Answer 1

We have, f(x) = `{{:(x^2/2",",  "if"  0 ≤ x ≤ 1),(2x^2 - 3x + 3/2",",  "if"  1 < x ≤ 2):}` at x = 1

At x = 1

L.H.L. = `lim_(x -> 1^-)  x^2/2`

= `lim_("h" -> 0) (1 - "h")^2/2`

= `lim_("h" -> 0) (1 + "h"^2 - 2"h")/2`

= `1/2`

 R.H.L. = `lim_(x -> 1^+) (2x^2 - 3x + 3/2)`

= `lim_("h" -> 0) [2(1 + "h")^2 - 3(1 + "h") + 3/2]`

= `2 - 3 + 3/2`

= `1/2`

Also f(1) = `1^2/2 = 1/2`

Thus L.H.L. = R.H.L. = f(1)

Hence, f(x) is continuous at x = 1.