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?

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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?

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Answer 1 · 2018-02-13T15:45:52

$h = \frac{1}{4}, k = \frac{17}{16}$ $h = 1/4, k= 17/16$ (assuming the function is $- {(x - h)}^{2} + k$ $-(x-h)^2+k$ for $x > 0$ $x>0$

Explanation:

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

This means that
$- h^{2} + k = 1$ $-h^2+k = 1$

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

So
$2 h = 0.5$ $2h = 0.5$

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

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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?

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