Modulus Question

21:31 Publicado por Mario Galarza

AppId is over the quota
AppId is over the quota
The function f is defined for all real values of x by
f(x) = 2 - (x+1)^\frac{1}{3}

The diagram shows this (it shows you intersections).

Find the set values of x for which f(x) = |f(x)|

Well, the root of the equation is x=7 and on the left side of this (you can see on the graph), all of the f(x) values are positive. On the right they're all negative. So it's clear for f(x) = |f(x)| that it'll be on the left of the root.

But my question is... Is the set of x values x < 7 or is it x \leq 7

Technically when x=7, f(x) = |f(x)|... 0 = |0|... But yeah, I always get confused about little things like this.

Last edited by TooEasy123; 30 Minutes Ago at 14:01.

View the original article here

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