Expand your knowledge base with the help of IDNLearn.com's extensive answer archive. Join our community to receive prompt and reliable responses to your questions from knowledgeable professionals.

[tex]\[
\begin{array}{l}
f(x) = 4x^2 + 7x - 3 \\
g(x) = 6x^3 - 7x^2 - 5
\end{array}
\][/tex]

Find [tex]\((f+g)(x)\)[/tex]:

A. [tex]\((f+g)(x) = 10x^3 - 8\)[/tex]

B. [tex]\((f+g)(x) = 6x^3 + 4x^2 - 8\)[/tex]

C. [tex]\((f+g)(x) = -6x^3 + 11x^2 + 7x + 2\)[/tex]

D. [tex]\((f+g)(x) = 6x^3 - 3x^2 + 7x - 8\)[/tex]


Sagot :

Let's find the resulting function, [tex]\( (f+g)(x) \)[/tex], by adding the given functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] step-by-step.

Given:
[tex]\[ f(x) = 4x^2 + 7x - 3 \][/tex]
[tex]\[ g(x) = 6x^3 - 7x^2 - 5 \][/tex]

To find [tex]\( (f+g)(x) \)[/tex], we add [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ (f+g)(x) = f(x) + g(x) \][/tex]
[tex]\[ (f+g)(x) = (4x^2 + 7x - 3) + (6x^3 - 7x^2 - 5) \][/tex]

Now, let's combine like terms:

1. [tex]\( x^3 \)[/tex] term:
Only [tex]\( g(x) \)[/tex] has a [tex]\( x^3 \)[/tex] term:
[tex]\[ 6x^3 \][/tex]

2. [tex]\( x^2 \)[/tex] term:
Combine the [tex]\( x^2 \)[/tex] terms from [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ 4x^2 - 7x^2 = -3x^2 \][/tex]

3. [tex]\( x \)[/tex] term:
Only [tex]\( f(x) \)[/tex] has an [tex]\( x \)[/tex] term:
[tex]\[ 7x \][/tex]

4. Constant term:
Combine the constants from [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ -3 - 5 = -8 \][/tex]

Now, combine all these terms together:
[tex]\[ (f+g)(x) = 6x^3 - 3x^2 + 7x - 8 \][/tex]

Thus, the resulting function is:
[tex]\[ (f+g)(x) = 6x^3 - 3x^2 + 7x - 8 \][/tex]

From the given options, this matches Option D.

Therefore, the correct answer is:
[tex]\[ \boxed{\text{D. } (f+g)(x)=6 x^3-3 x^2+7 x-8} \][/tex]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.