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June 15th, 2019, 10:46 AM   #1
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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.
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June 15th, 2019, 11:17 AM   #2
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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.
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June 15th, 2019, 11:22 AM   #3
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Math Focus: Yet to find out.
At your service m'lord

https://www.desmos.com/calculator/kiq4ivf6u5
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June 15th, 2019, 11:42 AM   #4
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Quote:
Originally Posted by Farzin View Post
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 $f \left(-\dfrac{b}{2a} \right) = f(3/2) = 5/4$
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June 15th, 2019, 11:58 AM   #5
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Thank you very much, that was a great help. Then I was right.
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