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Extreme values of function.

  • Thread starter Ofey
  • Start date
75
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1. Homework Statement

What are the extreme values for the function

[tex]f:f(x)=\int_0^x(|t|-|t-1|) \ d t [/tex]

in the interval [tex][-1,3][/tex]



2. Homework Equations

-



3. The Attempt at a Solution

I tried to simplify the function by getting rid of the absolute values.

This gave me three different possibilities depending on the value for t.


[tex]f:f(x)=\begin{cases} \int_0^x-1 \ d t, & \mbox{if } t \leq 0 \\ \int_0^x(2t-1) \ d t, & \mbox{if } 0<t<1 \\ \int_0^x1 \ d t ,& \mbox{if } t\geq1 \end{cases}[/tex]

Which I got to be

[tex]f:f(x)=\begin{cases} -x, & \mbox{if } t \leq 0 \\ x^2-x, & \mbox{if } 0<t<1 \\ x ,& \mbox{if } t\geq1 \end{cases}[/tex]

This is where it gets tricky for me. How do I treat the fact that the boundaries for the "different" functions are determined by t, when the variable is x. How to pursue the solution from this point?
 
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Using the Fundamental Theorem of Calculus, you could have noted that the integral is equivalent to F(x) - F(0) where F'(t) = |t| - |t - 1|. Thus, to find the extremal values, we just need to look at the behavior of the integrand: where it is undefined and where it is 0 and the values of the function at the endpoints.
 
75
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I managed to solve the problem using my method. However, it would be interesting if you could elaborate a bit more how you would have solved it. I understand that it is possible to use the integrand to find out the x-values which correspond to maximum/minimum values, but you still need the (proper) integral in which you "put" the x-values (which correspond to max/min values) to get the corresponding y-values for the minimum/maximum. So you eventually have to integrate no matter what?
 

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