Question

F(x)=e^0.5x when x<=0 and f(x)=-(x-h)2+k when x>0 find h and k such that f(x) is both continuous and smooth at x=0?

Answer 1

`h = 1/4, k= 17/16` (assuming the function is `-(x-h)^2+k` for `x>0`

Explanation:

For the function to be continuous, you need its left hand and right hand limits to both match its value at `x = 0`

This means that
`-h^2+k = 1`

To be smooth the left hand and right hand derivatives must match at `x=0`. Here the right hand derivative is that of `e^{0.5x}` at `x=0`, and this is equal to 0.5. The left hand derivative is that of `-(x-h)^2+k`, which is `-2(x-h)` and has a value `2h` at `x=0`

So
`2h = 0.5`

Thus `h= 0.25` and `k = 1-0.25^2=17/16`