Finding Extreme Values of a Function in a Given Interval

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SUMMARY

The extreme values of the function defined by the integral \( f(x) = \int_0^x (|t| - |t-1|) \, dt \) in the interval [-1, 3] can be determined by analyzing the piecewise function derived from the absolute values. The function simplifies to three cases: \( f(x) = -x \) for \( t \leq 0 \), \( f(x) = x^2 - x \) for \( 0 < t < 1 \), and \( f(x) = x \) for \( t \geq 1 \). To find the extreme values, one must evaluate the behavior of the integrand, identify critical points where it is zero or undefined, and compute the function values at the endpoints of the interval.

PREREQUISITES
  • Understanding of integral calculus, specifically the Fundamental Theorem of Calculus.
  • Familiarity with piecewise functions and their properties.
  • Knowledge of absolute value functions and their graphical interpretations.
  • Ability to find critical points and evaluate functions at specific intervals.
NEXT STEPS
  • Study the Fundamental Theorem of Calculus and its applications in finding extreme values.
  • Learn about piecewise function analysis and how to handle transitions between different cases.
  • Explore methods for finding critical points of functions, including derivatives and the first derivative test.
  • Investigate the properties of absolute value functions and their impact on integrals and derivatives.
USEFUL FOR

Students studying calculus, particularly those focusing on integration and optimization problems, as well as educators seeking to enhance their teaching methods for extreme value problems.

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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]



Homework Equations



-



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.
 
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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