# Thread: Express f(x) in terms of the Heaviside function

1. ## Express f(x) in terms of the Heaviside function

This question is attached.

Just in case you cannot see it, it asks to express the function f(x) in terms of the Heaviside function.

f(x)=x+1 when x<-1
3x when -1<x<1
2-x when x>1

I don't understand how to do this at all so please go easy on me. Baby steps .

2. ## Re: Express f(x) in terms of the Heaviside function

Originally Posted by softstyll
This question is attached.

Just in case you cannot see it, it asks to express the function f(x) in terms of the Heaviside function.

f(x)=x+1 when x<-1
3x when -1<x<1
2-x when x>1

I don't understand how to do this at all so please go easy on me. Baby steps .
It may help to look at f  '(x) . It is more directly related to the Heaviside function than is f(x).

Do you know what the Heaviside function is?

3. ## Re: Express f(x) in terms of the Heaviside function

For x< -1, f(x)= x+ 1. For -1< x< 1, f(x)= 3x= x+ 1+ (3x- (x+1))= x+1+ (2x- 1). For x> 1, f(x)= 2- x= 3x+ (2- x- 3x)= 3x+ (2- 4x). How use those with H(x+1) and H(x-1).

4. ## Re: Express f(x) in terms of the Heaviside function

Originally Posted by SammyS
It may help to look at f  '(x) . It is more directly related to the Heaviside function than is f(x).

Do you know what the Heaviside function is?
Yes I know the definition of the heaviside function

5. ## Re: Express f(x) in terms of the Heaviside function

Originally Posted by HallsofIvy
For x< -1, f(x)= x+ 1. For -1< x< 1, f(x)= 3x= x+ 1+ (3x- (x+1))= x+1+ (2x- 1). For x> 1, f(x)= 2- x= 3x+ (2- x- 3x)= 3x+ (2- 4x). How use those with H(x+1) and H(x-1).
How did you get those expressions for f(x)?

6. ## Re: Express f(x) in terms of the Heaviside function

Originally Posted by softstyll
How did you get those expressions for f(x)?
HoI merely used a bit of algebra.

7. ## Re: Express f(x) in terms of the Heaviside function

For example, if the problem had been
$f(x)= x^2$ for $x$< 2 and $f(x)= x^3- 3x$ for $x\ge 2$

I would have thought "on the left (x< 2) I have $x^2$ so I have to have that without any H. For x> 2, I want $x^2+ f(x)H(x)= x^2+ f(x)= x^3- 3x$ so $f(x)= x^3- 3x- x^2$, the formula wanted in the second interval with the formula from the first region subtracted off to cancel the $x^2$ that is no longer wanted.

In other words, $f(x)= x^2+ H(x-2)(x^3- 3x-x^2)$.