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At the moment, I am trying to work on a simple integral, involving an absolute value function. However, I am not just trying to merely solve it; I am undertaking to write, in detail, of everything I am doing.


So, the function is $f(x) = |x^2 + 3x - 4|$. I know that this isn't an algebraic-like function, so we can't evaluate it as one; but, by using the definition of absolute value, we can rewrite it as one.

The function $f(x)$, without the absolute value signs, can take on both positive and negative values; so, in order to retain the strictly positive output that the absolute value function demands, we have to put a negative sign in front of the algebraic definiton of our absolute vaue function on the interval where the values yield a negative value, so we'll get a double negative -(-), resulting in the positve we want.


This is how far I've gotten so far. From what i've been taught, in order to find the intervals where the function is positive and where it is negative, you have to find the values that make the function zero, and create test intervals from those values. For instance, the zeros of the function above are $x = -4$ and $x = 1$; our test intervals are then $(- \infty, -4)$, $(-4, 1)$, and $(1, \infty)$ My question is, why does finding the zeros of the function guarantee that we will find those precise test intervals?

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Hint: If $f$ is continuous, $f(c)>0$, and $f(d)<0$, then for some $e$ between $c$ and $d$, $f(e)=0$. So if you locate every zero of $f$ ... –  David Mitra Feb 22 '13 at 23:06
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Maybe work through Example 5 here to set the stage. –  Amzoti Feb 22 '13 at 23:06
    
Since you have the absolute value function, you want to simplify the computation by finding intervals where the function is strictly positive or negative. Since a continuous function can only change sign by passing through zero, this is why you are looking for zeros. –  copper.hat Feb 22 '13 at 23:06
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3 Answers 3

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Let us use your example to illustrate. If $x<-4$, then $x+4<0$ and $x-1<0$ so their product is positive. If $-4<x<1$, then $x+4>0$ and $x-1<0$ so their product is negative. If $x>1$, then $x+4>0$ and $x-1>0$ so their product is positive. The zeros help us split up the real line into these intervals where the individual factors are positive or negative so that we can determine if the product is positive or negative.

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Since polynomials and absolute value of continuous functions are continuous , the zeros are

the points for possible sign change.

Then you can write piecewisely your function and integrate..

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But the absolute value function isn't a polynomial... –  Mack Feb 23 '13 at 1:47
    
Oh, wait, do you perhaps mean the function without the absolute value bars? If that's the case, that's interesting. And that would explain why the graph $|x^2|$ looks the way it does. The portion of the graph of $x^2$ that is underneath the x-axis corresponds to negative outputs; by applying the absolute value function to $x^2$, that "cup" underneath the x-axis is inverted. –  Mack Feb 23 '13 at 1:57
    
i clarified above.. –  Halil Duru Feb 23 '13 at 2:07
    
well , $|x^2|=x^2.$ –  Halil Duru Feb 23 '13 at 2:10
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Continuous function does not change sign between two consecutive zeroes.

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Well, why is that true? –  Mack Feb 23 '13 at 1:47
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