Question

Limit - Piecewise Function
Three conditions given:
limit as x approaches of f(x) exists
limit as x approaches of f'(x) exists
f'(1) exists

Answer 1

2x^3-3 , x<1
6x-4, x>1

Answer 2

I know condition one is false because the limits are not the same -1 does not equal 2
I know condition two is true because limit of f'(x) is the same 6 = 6
I am not sure f'(1) exist, I am saying it does not since (1,6) does exist in the piecewise but I am not sure

Answer 3

@bahrom7893

Answer 4

to make sure that f'(1) doesn't exist, you can graph it and see if there's a cusp or something at x=1.

Answer 5

thanks