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 June 15th, 2019, 10:46 AM #1 Newbie   Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 Maximum/Minimum Please help me with this question: If a function $f(x)$ in the domain $x ∈ [0, 2]$ is $f(x) = |x − 1| + |x^2 − 2x|$, then the minimum value is $[1-8]$ and the maximum one is $[1-9]$ . My answer is the maximum is $\dfrac 5 4$ and the minimum is $1$ but I think I am wrong Last edited by Farzin; June 15th, 2019 at 11:23 AM.
 June 15th, 2019, 11:17 AM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,585 Thanks: 1430 first thing I would do is hit an online graphic site and graph the thing. The minimum does indeed appear to be 1, and the maximum is $\dfrac 5 4$ as well. I can't say I understand how you came up with these correct answers given what you've written. Thanks from topsquark
 June 15th, 2019, 11:22 AM #3 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,839 Thanks: 653 Math Focus: Yet to find out. At your service m'lord https://www.desmos.com/calculator/kiq4ivf6u5 Thanks from topsquark
June 15th, 2019, 11:42 AM   #4
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Quote:
 Originally Posted by Farzin Please help me with this question: If a function $f(x)$ in the domain $x ∈ [0, 2]$ is $f(x) = |x − 1| + |x^2 − 2x|$, then the minimum value is $[1-8]$ and the maximum one is $[1-9]$ . My answer is the maximum is $\dfrac 5 4$ and the minimum is $1$ but I think I am wrong
endpoint values ...

$f(0) = 1$, $f(2) = 1$

for $x \in (0,1)$, $f(x) = (1-x)+(2x-x^2) = 1 + x - x^2$

max on this interval is $f \left(-\dfrac{b}{2a} \right) = f(1/2) = 5/4$

for $x \in [1,2)$, $f(x) = (x-1)+(2x-x^2) = -1 + 3x - x^2$

$f(1) =1$, and max on this interval is $f \left(-\dfrac{b}{2a} \right) = f(3/2) = 5/4$

 June 15th, 2019, 11:58 AM #5 Newbie   Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 Thank you very much, that was a great help. Then I was right.

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