Question

Show that the function f(x) = `{(x^2, x<=1),(1/2, x>1):}` is continuous at x = 1 but not differentiable.

Answer 1

Continuity at x = 1

`(x = 1)= x^2 = (1)^2 = 1`

`lim_(x->1^+) f(x) = lim_(x->1^+) 1/x = 1`

`lim_(x->1^-) f(x) = lim_(x->1^-) x^2 = 1`

`:. f(x = 1) =  lim_(x->1^-) f(x) = lim_(x->1^+) f(x) = 1`

:. f(x) is continuous at x = 1

Now differentiable at x = 1

(R.H.D at x = 1) = `lim_(x->1^+) (f(x) - f(1))/(x -1)`

`=lim_(x->1) (1/x- 1)/(x - 1)`

`= lim_(x-> 1) (-(x-1) 1/x)/(x-1)`

`= = -1/1 = -1`

(L.H.D at x = 1) = `lim_(x -> 1^(-)) (f(x)-f(1))/(x-1)`

`= lim_(x->1^1) (x^2 -1)/(x - 1) = 2`

`:. l.H.D != R.H.D`

:. f(x) is not differentiable at x = 1