# 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?

Feb 13, 2018

$h = \frac{1}{4} , k = \frac{17}{16}$ (assuming the function is $- {\left(x - h\right)}^{2} + k$ 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 ${e}^{0.5 x}$ at $x = 0$, and this is equal to 0.5. The left hand derivative is that of $- {\left(x - h\right)}^{2} + k$, which is $- 2 \left(x - h\right)$ and has a value $2 h$ at $x = 0$

So
$2 h = 0.5$

Thus $h = 0.25$ and $k = 1 - {0.25}^{2} = \frac{17}{16}$