# Maximum/Minimum

#### Farzin

If a function $f(x)$ in the domain $x âˆˆ [0, 2]$ is
$f(x) = |x âˆ’ 1| + |x^2 âˆ’ 2x|$,
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:

#### romsek

Math Team
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.

• 1 person

#### Joppy

• 1 person

#### skeeter

Math Team
If a function $f(x)$ in the domain $x âˆˆ [0, 2]$ is
$f(x) = |x âˆ’ 1| + |x^2 âˆ’ 2x|$,
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 $f \left(-\dfrac{b}{2a} \right) = f(3/2) = 5/4$

• 1 person

#### Farzin

Thank you very much, that was a great help. Then I was right.