SOLUTION: Let f(X)=[-x-3 , x<-1 , cubic root of x , -1<=x<2 , x^2-x , x>=2].Find f(-1) , f(1) , f(2)

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Question 1197856: Let f(X)=[-x-3 , x<-1 , cubic root of x , -1<=x<2 , x^2-x , x>=2].Find f(-1) , f(1) , f(2)
Found 2 solutions by MathLover1, math_tutor2020:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Piecewise[{{ -x-3+, x%3C-1 }, { root%283%2Cx%29%7D%7D+%7D%2C%7B%7B%7B-1%3C=x%3C2 }, { x%5E2-x, x%3E=2 } }]

desmos-graph-15
Find
f(-1) ->use -x-3 ,because x%3C-1
f(1) ->use root%283%2Cx%29 ,because -1%3C=x%3C2
f(2) ->use x%5E2-x, because x%3E=2
f%28-1%29++=-%28-1%29-3=1-3=-2
f%281%29=root%283%2C1%29=1
f%282%29=2%5E2-2=4-2=2


Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

The tutor @MathLover1 is 2/3 correct.
f(1) and f(2) are correct, but f(-1) is not.

To compute f(-1), you should use f%28x%29+=+root%283%2Cx%29 since that applies for -1+%3C=+x+%3C+2 which is where x = -1 resides.

IF x+%3C+-1 was x+%3C=+-1, then you would use the first piece. However, we don't have the "or equal to" portion.