IDNLearn.com is designed to help you find the answers you need quickly and easily. Join our knowledgeable community and access a wealth of reliable answers to your most pressing questions.
Sagot :
To find [tex]\((f + g)(x)\)[/tex], we need to combine the functions [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex].
Given:
[tex]\[ f(x) = \frac{x}{2} - 2 \][/tex]
[tex]\[ g(x) = 2x^2 + x - 3 \][/tex]
Our goal is to add these two expressions together, that is,
[tex]\[ (f + g)(x) = f(x) + g(x) \][/tex]
Let's write out the functions with the same variable [tex]\(x\)[/tex]:
[tex]\[ (f + g)(x) = \left( \frac{x}{2} - 2 \right) + \left( 2x^2 + x - 3 \right) \][/tex]
To add these expressions, we combine like terms:
1. Combine the [tex]\(x^2\)[/tex] terms:
[tex]\[ 2x^2 \][/tex]
2. Combine the [tex]\(x\)[/tex] terms:
[tex]\[ \frac{x}{2} + x = \frac{x}{2} + \frac{2x}{2} = \frac{3x}{2} \][/tex]
3. Combine the constant terms:
[tex]\[ -2 - 3 = -5 \][/tex]
Putting these together, we get:
[tex]\[ 2x^2 + \frac{3x}{2} - 5 \][/tex]
So, the resulting function [tex]\((f + g)(x)\)[/tex] is:
[tex]\[ (f + g)(x) = 2x^2 + \frac{3x}{2} - 5 \][/tex]
This is our final expression.
Given:
[tex]\[ f(x) = \frac{x}{2} - 2 \][/tex]
[tex]\[ g(x) = 2x^2 + x - 3 \][/tex]
Our goal is to add these two expressions together, that is,
[tex]\[ (f + g)(x) = f(x) + g(x) \][/tex]
Let's write out the functions with the same variable [tex]\(x\)[/tex]:
[tex]\[ (f + g)(x) = \left( \frac{x}{2} - 2 \right) + \left( 2x^2 + x - 3 \right) \][/tex]
To add these expressions, we combine like terms:
1. Combine the [tex]\(x^2\)[/tex] terms:
[tex]\[ 2x^2 \][/tex]
2. Combine the [tex]\(x\)[/tex] terms:
[tex]\[ \frac{x}{2} + x = \frac{x}{2} + \frac{2x}{2} = \frac{3x}{2} \][/tex]
3. Combine the constant terms:
[tex]\[ -2 - 3 = -5 \][/tex]
Putting these together, we get:
[tex]\[ 2x^2 + \frac{3x}{2} - 5 \][/tex]
So, the resulting function [tex]\((f + g)(x)\)[/tex] is:
[tex]\[ (f + g)(x) = 2x^2 + \frac{3x}{2} - 5 \][/tex]
This is our final expression.
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.