Write the absolute value function as a piecewise-defined function with linear parts.

Mathematics

1 answer:

KengaRu [80]2 years ago

6 0

The piecewise-defined function with linear parts f(x)= |2x + 9|

f(x) = {2x + 9 x ≥ -9/2)

{-2x - 9 x < 9/2}

<h3>How to write piecewise functions?</h3>

To write this function as a piecewise function, you have to use the definition of absolute value.

The absolute value equation is; f(x)= |2x + 9|

The definition says that;

|x| = {x x ≥ 0)

{-x x < 0}

Applying this definition to the function you get:

|2x + 9| = {2x + 9 2x + 9 ≥ 0)

{-2x - 9 2x + 9 < 0}

After simplifying the inequalities you get:

|2x + 9| = {2x + 9 x ≥ -9/2)

{-2x - 9 x < 9/2}

So finally you can write the piecewise function as:

f(x) = {2x + 9 x ≥ -9/2)

{-2x - 9 x < 9/2}

Read more about piecewise function at; brainly.com/question/18499561

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Which of the following ordered pairs belongs to the graph of f(x)

Lera25 [3.4K]

Answer:

The second choice, $(-1,\, 12)$ .

Step-by-step explanation:

Note, that the expression $y = f(x)$ is an equation. A point $(x,\, y)$ is on the graph of $y = f(x)\!$ if and only if the value of $x$ and $y$ satisfy this equation; that is: in other words, the $y\!$ -coordinate of that point (the second number in the tuple) should be equal to $f(x)$ , which is equal to $(-3\, x +9)$ (evaluated where $x\!$ is equal to the first number in the tuple.

For each tuple in the choices, calculate the value of $f(x)$ where $x$ is equal to the first number of each tuple. Compare the result to the second number in that tuple. That choice corresponds to a valid point on $y = f(x)$ only if these two numbers match.

First choice: $x = 1$ , $f(x) = f(1) = -3 + 9 = 6$ . That's not the same as the second number, $-10$ . Therefore, this point isn't on the graph of $y = f(x)$ .
Second choice: $x = -1$ , $f(x) = f(-1) = (-3)\times (-1) + 9 = 3 + 9 = 12$ . That matches the second number in the tuple. Therefore, this point is on the graph of $y = f(x)$ .
Third choice: $x = 2$ , $f(x) = f(2) = (-3)\times 2 + 9 = 3$ . That's not the same as the second number, $(-3)$ . Therefore, this point isn't on the graph of $y = f(x)$ .
Fourth choice: $x = -2$ , $f(x) = f(-2) = (-3)\times (-2) + 9 = 15$ . That's not the same as the second number, $13$ . Therefore, this point isn't on the graph of $y = f(x)$ .