What is a piecewise-defined function and how do I write one?

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SUMMARY

A piecewise-defined function is expressed using the Heaviside step function, H(x), which is crucial for defining different expressions based on the value of x. In the provided example, f(x) = x(1-x)H(x-1) + (4x-x^2-4)H(x-2), the function evaluates to 0 for x < 1, x(1-x) for 1 ≤ x < 2, and 3x - 4 for x ≥ 2. The Heaviside function H(x) is defined as 0 for x < 0 and 1 for x ≥ 0, allowing for the segmentation of the function into distinct cases. The final expression can be simplified to f(x) = {0, x(1-x), 3x-4} depending on the range of x.

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hi, i missed a lecture on this topic and i am really lost...there is a question that asks to write a function in the piecewise-defined form. i have no idea what that means and its not in the textbook of the course...looked everywhere online and can't find anything on it...

they give the function in the form of f(x)=x(1-x)H(x-1)+(4x-x^2-4)H(x-2)

have no idea how to proceed...dont understand what H is doing in there...can someone provide some explanation and a little insight into this topic...would be appreciated.
 
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You might want to check your textbook for the definition of H(x), the Heaviside step function.

H(x) is defined as "0 is x< 0, 1 if x\ge 0".

From that H(x-1) is 0 if x< 1, 1 if x\ge 1 and H(x-2) is 0 if x< 2, 1 if x\ge 2.

In this example, if x< 1, then both H(x-1) and H(x-2) are 0. If x< 1, f(x)x(1-x)(0)+ (4x- x2-4)(0)= 0. If 1\le x&lt; 2 then H(x-1)= 1 while H(x-2) is 0. f(x)= x(1-x)(1)+ (4x- x2-4)(0)= x(1-x). If x\ge 2 both H(x-1) and H(x-2) are 1 so f(x)= x(1-x)(1)+ (4x- x2-4)(1)= x(x-1)+ 4x- x2- 4= x- x2+ 4x- x2- 4= 3x- 4.

That could also be written
f(x)= \left[\begin{array}{c} 0 \\ x(1-x) \\ 3x- 4\end{array}\right
 
thanks, i understand...but for the third part...isn't it -2x^2+5x-4? i think u made a small error in the final calculation...
 

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