Question

when i have something like: f(x) = |x^2-x| what does it mean?

Answer 1

this is the absolute value symbol. What does the directions ask of you?

Answer 2

it means that the y value of the ordered pair is x^2-x and that the answer will be 0 or positive.

Answer 3

well it's asking me to find the derivative of that function above. so i'm think there are more than 1 derivatives to find. for instance derivative of f(x) when x >= 0 an derivative when x <0 but i'm not so sure

Answer 4

you could write it as
for x≥ 1 f(x)= x^2-x
for 0≤x ≤1 f(x)= x-x^2

for x ≤ 0 f(x)= x^2+x

if I did that correctly...

Answer 5

for x <0 x^2-x is much better

Answer 6

Remember, the definition of |x| is:

|x| = x, if x >= 0
|x| = -x, if x < 0.

So: f(x) = |x^2-x| means:

f(x)= x² - x, if x² - x >= 0
f(x)= -x² + x, if x² - x <0

So, to know which of the two formulas applies for certain values of x, you will have to solve x² - x = 0.
It has two (simple to find) solutions. Put them on a number line and then make a sign scheme of x² - x. Then you know which of the two formulas applies where...

Answer 7

well i have this: http://www.sosmath.com/calculus/diff/der13/img5.gif which i don't understand at all! how the get the if's and how they change the signs.

Answer 8

if x > 1 then x^2 is bigger than x and x^2-x is positive | positive #| is just the #
if x between 0 and 1 x^2 is less than x and x^2 -x is negative
the absolute value operation changes this to positive: you could say it multiplies by -1:
-(x^2-x)

Answer 9

The sign scheme is (see image).

So for 0 < x < 1 you have:
f(x) = -(x² - x) = -x² + x (because x²-x is negative there)

for x <=0 and for x >=1 you have:
f(x) = x² - x (because x² - x is positive there, so you just remove the |..| )