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2 years ago

How do you determine the domain of a function ?

Mathematics

2 answers:

Sunny_sXe [5.5K]2 years ago

8 0

Answer:

Let y = f(x) be a function with an independent variable x and a dependent variable y.

If a function f provides a way to successfully produce a single value y using for that purpose a value for x then that

chosen x-value is said to belong to the domain of f. If there is a requirement that a y-value produced by a function

must be a real number, the following conditions are commonly checked:

1. Denominators cannot equal 0.

2. Radicands (expressions under a radical symbol) of even roots (square roots, etc)

cannot have a negative value.

3. Logarithms can only be taken of positive values.

4. In word problems physical or other real-life restrictions might be imposed, e.g. time is

nonnegative, number of items is a nonnegative integer, etc.

stiks02 [169]2 years ago

5 0

Answer:

the domain is a set of all possible numbers inputs for that function, THe range is all the possible output numbers

i hope this helps

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Write a system of linear inequalities so the points (1, 2) and (4, -3) are solutions of the system, but the point (-2, 8) is not

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Answer: $\left \{ {{y\leq -\frac{5}{3}x + \frac{11}{3}} \atop {y < 8}} \right$

<u>Step-by-step explanation:</u>

(1, 2) and (4, -3)

First, find the slope (m): $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

m = $\frac{-3-2}{4-1}$

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y - 2 = $-\frac{5}{3}$ (x - 1)

y - 2 = $-\frac{5}{3}x$ + $\frac{5}{3}$

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Then, determine which inequality symbol will result in (-2, 8) being False (since it is not a solution):

y ___ $-\frac{5}{3}x$ + $\frac{11}{3}$

8 ___ $-\frac{5}{3}(-2)$ + $\frac{11}{3}$

8 ___ $\frac{1}{3}$

8 > $\frac{1}{3}$

so ≤ makes the statement False

⇒ y ≤ $-\frac{5}{3}x$ + $\frac{11}{3}$

****************************************************************

If you need a "system" of inequalities, then you need another equation.

There are infinite possibilities. Graph the points to see what works.

y < 8 or x > -2 are two examples.

---> I just realized that the system can be simpler than what I did above:

$\left \{ {{x>-2} \atop {y < 8}} \right$

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