If f(x) = 2x-1 and g(x) = x^2-3x -2, find (f + g) (x).

Question

If f(x) = 2x-1 and g(x) = x^2-3x -2, find (f + g) (x).

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3 answers
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Answer 1

Bot GPT 4-o mini answered
7 hours ago

To find \((f + g)(x)\), we first need to define what it means:

\[
(f + g)(x) = f(x) + g(x).
\]

Given:
- \(f(x) = 2x - 1\)
- \(g(x) = x^2 - 3x - 2\)

Now, we will add \(f(x)\) and \(g(x)\):

\[
(f + g)(x) = (2x - 1) + (x^2 - 3x - 2).
\]

Now, we'll simplify this expression:

1. Combine like terms:
\[
(2x - 1) + (x^2 - 3x - 2) = x^2 + (2x - 3x) + (-1 - 2).
\]

2. This simplifies to:
\[
x^2 - x - 3.
\]

Therefore, the result is:

\[
(f + g)(x) = x^2 - x - 3.
\]

Answer 2

anonymous answered
7 hours ago

Does this table represent a function? Why or why not?

X. Y.
2. 1
4. 3
6. 3
8. 4
10. 6
Answer:
=---
" Answer 1-----.
:: Yes, because every x-value corresponds to exactly one y-value.
= No, because one x-value corresponds to two different y-values.
:: Yes, because all of the x-values correspond to one or more y-values.
Previolis
》
¿ No, because two of the y-values are the same.

Answer 3

Bot GPT 4-o mini answered
7 hours ago

To determine whether the given table represents a function, we need to apply the definition of a function. A relation is considered a function if every input (or x-value) corresponds to exactly one output (or y-value).

Here is the provided data:

| X | Y |
|----|----|
| 2 | 1 |
| 4 | 3 |
| 6 | 3 |
| 8 | 4 |
| 10 | 6 |

Now, let's analyze the x-values:

- \( x = 2 \) maps to \( y = 1 \)
- \( x = 4 \) maps to \( y = 3 \)
- \( x = 6 \) maps to \( y = 3 \)
- \( x = 8 \) maps to \( y = 4 \)
- \( x = 10 \) maps to \( y = 6 \)

In this case:

- Each of the x-values (2, 4, 6, 8, 10) has a corresponding y-value.
- The x-value of \( 6 \) corresponds to \( y = 3 \), which is already associated with \( x = 4 \), but that does not disqualify it from being a function because the same y-value can be associated with different x-values.

Since every x-value has exactly one y-value associated with it, the answer is:

**Yes, because every x-value corresponds to exactly one y-value.**