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tomiko
when i have something like: f(x) = |x^2-x| what does it mean?
this is the absolute value symbol. What does the directions ask of you?
it means that the y value of the ordered pair is x^2-x and that the answer will be 0 or positive.
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
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...
for x <0 x^2-x is much better
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...
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.
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)
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 |..| )