write the absolute value function as a piecewise function. y=I4x+1I+2x-3
We need the 'turning' point of your absolute value function, that is, the value of x which would make the thing inside the bars || equal to zero. That said, when is 4x + 1 equal to zero?
@terenzreignz when x=-1/4 ?
Precisely :) So that's going to be the turning point of your function :D \[\Large f(x) = \left\{\begin{matrix}&\qquad x \le -\frac14\\\\\\\\\\&\qquad x> -\frac14\end{matrix}\right.\]
Now, when x is less than -1/4 4x+1 is positive or negative?
@terenzreignz negative?
Yes. So, since it's negative, and in absolute value, it becomes positive when x is less than -1/4 Then we multiply it by -1 and it becomes -4x - 1 okay?
ok that makes sense so far
Right? Because |4x + 1| is basically the same as -4x - 1 whenever x is less than -1/4
lol okay, so when x is less than or equal to -1/4, then the thing inside the absolute value gets multiplied by -1 (to turn it positive) \[\Large f(x) = \left\{\begin{matrix}\color{blue}{-4x-1}+2x -3&\qquad x \le -\frac14\\\\\\\\\\&\qquad x> -\frac14\end{matrix}\right.\] We will simplify later. Anyway, when x is greater than -1/4 then |4x + 1| is positive, yes?
Rather, sorry (since |4x+1| is always positive anyway) when x is greater than -1/4 then 4x + 1 is positive, right?
oh ok!!! so f(x)={ 6x-2 when x greater than or equal to -1/4 -2x-4 when x < -1/4
@terenzreignz
Someone's eager :) LOL Correct :) \[\Large f(x) = \left\{\begin{matrix}-2x-4&\qquad x \le -\frac14\\\\\\\\\\6x-2&\qquad x> -\frac14\end{matrix}\right.\] Note that it doesn't matter if we switch the "or equal to" like this \[\Large f(x) = \left\{\begin{matrix}-2x-4&\qquad x \color{red}< -\frac14\\\\\\\\\\6x-2&\qquad x\color{red}{\ge}-\frac14\end{matrix}\right.\] Since both 'pieces' of the function have the same value when x = -1/4 Great job, by the way ^_^
ok thank you so much! i finally understand this stuff now! :)
Glad to hear it :) Signing off for now... ------------------------------ Terence out
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