# Determine whether f in f(x) = x^2 + 2x, x ≤ 1 and 3x, x > 1 is differentiable at x=1

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### 1 Answer

You first need to check if the function is continuous at x=1 such that:

`lim_(x->1,x<1) f(x) = lim_(x->,x>1) f(x) = f(1)`

`lim_(x->1,x<1) x^2 + 2x = 1 + 2*1 = 3`

`lim_(x->,x>1) 3x = 3*1 = 3`

`f(1) = 1+2*1 = 3`

Since `lim_(x->1,x<1) f(x) = lim_(x->,x>1) f(x) = f(1) = 3` , the function is continuous at x=1.

Since the function is continuous, you may check if the function is differentiable at x=1 such that:

`f'(1) = lim_(x->1,x<1) (x^2 + 2x - 3)/(x-1) `

`f'(1) = lim_(x->1,x<1) ((x-1)(x+3))/(x-1) = lim_(x->1,x<1) (x+3)`

`f'(1) = 1 + 3 = 4`

`f'(1) = lim_(x->1,x>1) (3x - 3)/(x - 1) = lim_(x->1,x>1) (3(x-1))/(x-1) = 3`

**Since `lim_(x->1,x<1) (x^2 + 2x - 3)/(x-1) = 4 != 3 = lim_(x->1,x>1) (3x - 3)/(x - 1)` , hence, the function is not differentiable at x = 1.**