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The absolute value function can be defined using piecewise notation.
A(z) =
(
2, 20
-X, I <0
Use this notation to find the following values:
1. A(10)
2. A(0)
3. A(-3)
4. A(3.14159)
5. A(x) = 7
6. A(x) = -5




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The Absolute Value Function Can Be Defined Using Piecewise Notation Az 2 20 X I Lt0 Use This Notation To Find The Following Values 1 A10 2 A0 3 A3 4 A314159 5 A class=

Sagot :

Answers:

  1. A(10) = 10
  2. A(0) = 0
  3. A(-3) = 3
  4. A(3.14159) = 3.14159
  5. A(x) = 7 leads to either x = 7 or x = -7
  6. A(x) = -5 has no solutions

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Explanations:

The piecewise function has two identities based on what the x input is.

If x = 0 or larger, then A(x) = x based on the top row.

Or, if x < 0, then A(x) = -x based on the second row.

So for an input like x = 10, we have A(10) = 10. The input is identical to the output. The same goes for x = 0 and x = 3.14159

Each output tells us how far away the input is from zero on the number line.

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For a negative input, we'll use the second row

A(x) = -x

A(x) = -(x)

A(-3) = -(-3)

A(-3) = 3

Showing that the number -3 is exactly 3 units away from zero on the number line. In other words, |-3| = 3.

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To solve A(x) = 7, we have to think what input(s) will lead to an output of 7.

What two numbers are 7 units away from zero on the number line? That would be -7 and 7.

If you plugged x = 7 into the piecewise function, you'll use the top row to get A(7) = 7. The bottom row will have A(-7) = 7.

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

There are no solutions to A(x) = -5 because the result of an absolute value is never negative. Negative distances do not make sense, so that's why absolute value is defined this way.

If you tried x = 5 then A(5) = 5

Trying x = -5 leads to A(-5) = 5

There's no way to get a negative output.

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