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JuanitaM
 2 years ago
Best ResponseYou've already chosen the best response.0this is the absolute value symbol. What does the directions ask of you?

Mertsj
 2 years ago
Best ResponseYou've already chosen the best response.0it means that the y value of the ordered pair is x^2x and that the answer will be 0 or positive.

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0well 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

phi
 2 years ago
Best ResponseYou've already chosen the best response.0you could write it as for x≥ 1 f(x)= x^2x for 0≤x ≤1 f(x)= xx^2 for x ≤ 0 f(x)= x^2+x if I did that correctly...

phi
 2 years ago
Best ResponseYou've already chosen the best response.0for x <0 x^2x is much better

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0Remember, the definition of x is: x = x, if x >= 0 x = x, if x < 0. So: f(x) = x^2x 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...

tomiko
 2 years ago
Best ResponseYou've already chosen the best response.0well 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.

phi
 2 years ago
Best ResponseYou've already chosen the best response.0if x > 1 then x^2 is bigger than x and x^2x 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^2x)

ZeHanz
 2 years ago
Best ResponseYou've already chosen the best response.0The sign scheme is (see image). So for 0 < x < 1 you have: f(x) = (x²  x) = x² + x (because x²x 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 .. )
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