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Vlad1618 [11]
2 years ago
14

Find the domain and range of the function. ​f(x)=24+x^2

Mathematics
1 answer:
masya89 [10]2 years ago
6 0

<h2>Domain and Range of a Function</h2><h3>Answer:</h3>

Domain: x \in \mathbb{R}

Range: y \in \{ \mathbb{R} | y \geqslant 24 \}

<h3>Step-by-step explanation:</h3>

Domain:

In finding the Domain of a function, the values for the input of the function should not make the output of the function <em>undefined</em> or <em>complex</em>. Because of this, we can think of the values for the input that make the output of the function <em>undefined</em> or <em>complex</em> so that we will <u>not include them in our Domain</u>. We can only make the output <em>undefined</em> if the input makes the denominator 0. In f(x), there's no value for x that makes the denominator 0 as it is constant, 1 (Note: All expressions implicitly have 1 as their denominator even though it's not written). We can only make the output of the function <em>complex</em> if the value of x makes the function take the n\text{th} root of a negative number where n \in \mathbb{E}. There's no radical sign in f(x) so we shouldn't worry about the output of f(x) being <em>complex</em>. Because there's no value for the input, x, that can make the output of f(x) <em>undefined</em> or <em>complex</em>, its Domain can be any number.

Domain: x \in \{\mathbb{R}\}

Range:

In finding the Range, it is actually the same logic as finding the Domain but first, we'll have to do a bit of rewriting for the given function.

Let y = f(x) so y = 24 +x^2.

First, we need to make x the subject of the equation.

Making x the subject:

y = 24 +x^2 \\ y -24 = 24 +x^2 -24 \\ y -24 = x^2 \\ \pm \sqrt{y -24} = \sqrt{x^2} \\ x = \pm \sqrt{y -24}.

In the Domain, we'll have to think of the value for the input, x, that makes the output <em>undefined</em> or <em>complex</em>. In the Range, the same logic for the Domain, we'll have to think for the value of y, that makes the x <em>undefined</em> or <em>complex</em>. That's why we made x in the equation the subject. In our rewritten equation, we can see that y -24 is under the <em><u>square</u></em> root. Which means if y -24 is negative, x will be <em>complex</em>. So we have to make y -24 be greater than or equal to 0 (y -24 \geqslant 0) so that x won't be complex.

Solving for the inequality, y -24 \geqslant 0

y -24 \geqslant 0 \\ y -24 +24 \geqslant 0 +24 \\ y \geqslant 24

Range: y \in \{ \mathbb{R} | y \geqslant 24 \}

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