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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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
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 |..| )