# continous piece wise function

• April 28th 2010, 10:03 PM
ffuh2
continous piece wise function
Create a continuous function f(x) on the interval [-4, 4] by drawing its graph which meets these conditions; f (0) = 2,
f '(2) = 0, f '(-2) {is undefined, f "(0) = 0, f "(-1) is undefined. There is a point of inflection when x = 1 and a minimum at x = -2/3. The limit of f(x) as x approaches 4 is -2.

Ok. It seems pretty easy to do a continous piece wise function. I could have a two linear function from some point between greater than 1 to 4 and -2 to -1, then create a function that is a constant from 4 to infinity and from -2 to negative infinity. I could then create a sin function from -1 to some point greater than 1 This would creat sharp corners at -2, -1 which means the function would be continuous but not differentiable at those points. I could manipulate the period to have the sin function with a derivative of zero at x = 0, f'(0) & f"(0) both equal zero. I could manipulate the a & b of a sine function to have the inflection at 1 and minimum at -2/3 and a limit of -2 if I set y = -2 at x => 4. My question Is there a way to do this numerically without creating a piecewise function? If so, how?
• April 29th 2010, 10:30 PM
ffuh2
Any thoughts on this one?
• April 30th 2010, 03:15 AM
HallsofIvy
Since you have points where the derivative of second derivative are undefined, try a rational function. Since you are to have f(2) and f'(2) equal to 0, have a factor of $(x- 2)^2$ in the numerator. It says f'(-2) is to be undefined but doesn't say anything about f(-2) so I would interpret that as meaning that f(-2) is defined. Put a factor of $(x+ 2)^2$ in the denominator and in the numerator but define the function at x= -1 as the value given by the other terms, etc.
• April 30th 2010, 08:42 AM
ffuh2
the only problem with that is that the function will not be continuous. The first and second derivative need to be undefined at certain points but the underlying function must me continous on all intervals. If I have a factor of (x + 2) in the denominator, it will go to infinity as x approaches -2.