00:01
Alright, so question 43 gives us a compound function, x squared plus 1, where x is less than equal to 1, and 2x, where x is greater than 1, and we're supposed to show that it's continuous and differentiable at x equals 1.
00:16
So first off, let's draw the graph.
00:18
So for everything x is less than 1, our graph is y equals 2x.
00:23
So that happens at 2, 1 with that line.
00:28
And then for anything greater than 1, the function has, the definition of x squared plus 1.
00:35
So x squared plus 1 when x equals 1 is 2.
00:41
Now since the limit as x approaches 1 from the positive side of f of x is equal to the limit as x approaches 1 from the negative side of f of x and all of that is equal to f of 1 itself common the function is continuous...