How to Plot Heaviside Function with x^2-1?

[radiator]

Hello,

I am not where this question goes, its not part of a homework either!
I am trying to figure out how to plot the heaviside (unit step) with such an expression
H(x^2-1)
so I do this(x^2-1) = 1 for x^2-1>0 -> x>+- 1
and H(x^2-1) = 0 for x^2-1<-0 -> x<-+1

But this tells me only that it equals 1 when it is x>1 because the rest would be zero! any clarification would be appreciated

 

[pasmith]

Hello,

I am not where this question goes, its not part of a homework either!
I am trying to figure out how to plot the heaviside (unit step) with such an expression
H(x^2-1)
so I do this(x^2-1) = 1 for x^2-1>0 -> x>+- 1
and H(x^2-1) = 0 for x^2-1<-0 -> x<-+1

But this tells me only that it equals 1 when it is x>1 because the rest would be zero! any clarification would be appreciated

If x^2 - 1 &gt; 0 then either x &lt; -1 or x &gt; 1.

 

[radiator]

so for x^2-1>0 I have
H(x^2-1) = 1 for x>1 and x>-1 ( so its equal to 1 from -1 to infinity)
and
H(x^2-1) = 0 for x<1 and x<-1 ( so its equal to 0 from 1 to negative infinity)

so why would I choose only : equals 1 for x>1 and x<-1, what about the other conditions?

 

[Mark44]

so for x^2-1>0 I have
H(x^2-1) = 1 for x>1 and x>-1 ( so its equal to 1 from -1 to infinity)
and
H(x^2-1) = 0 for x<1 and x<-1 ( so its equal to 0 from 1 to negative infinity)

so why would I choose only : equals 1 for x>1 and x<-1, what about the other conditions?

Let's back up a bit, since you're confused about the solution to x² - 1 > 0.
x² - 1 ≥ 0 ==> x ≥ 1 or x ≤ -1.
So H(x² - 1) = 1 for x ≥ 1 or x ≤ -1.

 

[radiator]

Thanks Mark,
I think I am missing a basic principle here about the ≥ relations

if I have a x^2 - 1 \geq 0 then solving for x is
x^2 - 1 = 0 \rightarrow x = \pm 1
so in the case of positive one
x\geq 1 and for negative one it changes to
x \leq -1 and the positive it remains the same, is that a property of the relation?
also why do we choose ≥ instead of just >

 

[Mark44]

Thanks Mark,
I think I am missing a basic principle here about the ≥ relations

Yes, I think so as well.

if I have a x^2 - 1 \geq 0 then solving for x is
x^2 - 1 = 0 \rightarrow x = \pm 1

There's a big difference between solving an equation (as above) and solving an inequality.
What you've written below doesn't make any sense to me.

so in the case of positive one
x\geq 1 and for negative one it changes to
x \leq -1 and the positive it remains the same, is that a property of the relation?
also why do we choose ≥ instead of just >

Because H(x) = 1 if x ≥ 0 and H(x) = 0 for x < 0. That's why.

For your function, H(x² - 1) = 1 for x² - 1 ≥ 0, so we need to find the intervals for which x² - 1 ≥ 0.

The part on the left side of the inequality equals 0 for x = -1 or x = 1. These two numbers divide the number line into three intervals: (-∞, -1), (-1, 1), and (1, ∞). By taking any number in each interval you can verify that the solution to x² - 1 ≥ 0 is (-∞, -1] U [1, ∞). This is the same as saying x ≤ -1 or x ≥ 1.

It would be helpful for you to review the technique of solving inequalities, especially quadratic inequalities.

 

[radiator]

Thanks so much Mark, this clarified it