Question 1197822
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Some of what you typed in is a bit garbled. But I'm assuming this is what the piecewise function should look like
*[illustration Screenshot_113.png]
If my assumption is incorrect, then please let me know.


That piecewise function breaks down into 3 cases.
Case A: If {{{x < -1}}} then {{{f(x) = -x-3}}}
OR
Case B: If {{{-1<=x < 2}}} then {{{f(x) = root(3,x)}}}
OR
Case C: If {{{x>=2}}} then {{{f(x) = x^2-x}}}
We can only pick exactly one case. This is because each interval mentioned has no overlap with any other. 
There's no way, for instance, to have x be something smaller than -1 and larger than 2 at the same time.


To compute f(-1) we plug in x = -1
The input x = -1 fits the description of case B since -1 is in the interval {{{-1<=x<2}}}
I.e. {{{-1 <= -1<2}}} is a true statement.


We'll plug x = -1 into the second piece.
{{{f(x) = root(3,x)}}}


{{{f(-1) = root(3,-1)}}}


{{{f(-1) = root(3,(-1)^3)}}}


{{{f(-1) = -1}}}


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Now let's compute f(1).


The input x = 1 fits the interval {{{-1<=x<2}}} so we'll go for case B again.


{{{f(x) = root(3,x)}}}


{{{f(1) = root(3,1)}}}


{{{f(1) = root(3,1^3)}}}


{{{f(1) = 1}}}


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Lastly let's find f(2).
The input x = 2 fits the interval {{{x >= 2}}}.
We'll go for case C this time.


{{{f(x) = x^2-x}}}


{{{f(2) = 2^2-2}}}


{{{f(2) = 2}}}


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<font color=red>Summary:</font>
f(-1) = -1
f(1) = 1
f(2) = 2
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