Show that f(x)=...?

Question

x^2+2, x≤1
x+2, x>1

is continuous but not differentiable @ x=1

please explain problem & graph

Answer 1

If x=1

x^2 +2 = 1^2 + 2 = 3
1 + 2 = 3

If x≤1 d/dx (x^2 + 2) = 2x
Plug x=1 and get 2x=2

If x>1 d/dx (x + 2) = 1

We have 2 values of f'(x) when x=1, thus f(x) is not differentiable at x = 1

Answer 2

At x = 1, f(x) = 1^2 + 2 = 3

As x tends to 1 from above, f(x) tends to 3 from above

So f(x) is continuous

f'(x) = 2x x≤1
= 1 x> 1

So as x tends to 1 from below, f'(x) tends to 2 from below
and as x tends to 1 from above, f'(x) is 1

Also, a f(1), it is impossible to draw a line tangential to the curve.

Answer 3

there will probably be a discontinuous point at x=1....so its not differentiable...and ust graph it on a calculator or something and only show the lines at x<1 and x>1