SOLUTION: Let f(x)=[-x-3 , cubic root of x, x^2-x, x < -1, -1 < or equal x < 2, x > or equal 2 Find f(-1), f(1), and f(2)

Algebra ->  Functions -> SOLUTION: Let f(x)=[-x-3 , cubic root of x, x^2-x, x < -1, -1 < or equal x < 2, x > or equal 2 Find f(-1), f(1), and f(2)      Log On


   

Question 1197822: Let f(x)=[-x-3 , cubic root of x, x^2-x, x < -1, -1 < or equal x < 2, x > or equal 2
Find f(-1), f(1), and f(2)
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Some of what you typed in is a bit garbled. But I'm assuming this is what the piecewise function should look like

If my assumption is incorrect, then please let me know.

That piecewise function breaks down into 3 cases.
Case A: If x+%3C+-1 then f%28x%29+=+-x-3
OR
Case B: If -1%3C=x+%3C+2 then f%28x%29+=+root%283%2Cx%29
OR
Case C: If x%3E=2 then f%28x%29+=+x%5E2-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%3C=x%3C2
I.e. -1+%3C=+-1%3C2 is a true statement.

We'll plug x = -1 into the second piece.
f%28x%29+=+root%283%2Cx%29

f%28-1%29+=+root%283%2C-1%29

f%28-1%29+=+root%283%2C%28-1%29%5E3%29

f%28-1%29+=+-1

-----------------------------------------

Now let's compute f(1).

The input x = 1 fits the interval -1%3C=x%3C2 so we'll go for case B again.

f%28x%29+=+root%283%2Cx%29

f%281%29+=+root%283%2C1%29

f%281%29+=+root%283%2C1%5E3%29

f%281%29+=+1

-----------------------------------------

Lastly let's find f(2).
The input x = 2 fits the interval x+%3E=+2.
We'll go for case C this time.

f%28x%29+=+x%5E2-x

f%282%29+=+2%5E2-2

f%282%29+=+2

-----------------------------------------

Summary:
f(-1) = -1
f(1) = 1
f(2) = 2