Alright, let's solve the problem step-by-step.
We are given two functions:
[tex]\[ f(x) = x^2 + 1 \][/tex]
[tex]\[ g(x) = -3x - 5 \][/tex]
We need to find the function [tex]\((f + g)(x)\)[/tex].
1. Find [tex]\((f + g)(x)\)[/tex]:
To find [tex]\((f + g)(x)\)[/tex], we simply add [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]:
[tex]\[ (f + g)(x) = f(x) + g(x) \][/tex]
Substitute the given functions into this expression:
[tex]\[ (f + g)(x) = (x^2 + 1) + (-3x - 5) \][/tex]
Combine like terms:
[tex]\[ (f + g)(x) = x^2 - 3x + 1 - 5 \][/tex]
[tex]\[ (f + g)(x) = x^2 - 3x - 4 \][/tex]
So, the function [tex]\( (f + g)(x) \)[/tex] is:
[tex]\[ (f + g)(x) = x^2 - 3x - 4 \][/tex]
2. Determine the domain of [tex]\((f + g)(x)\)[/tex]:
The domain of a function is the set of all possible input values (x-values) for which the function is defined. For a polynomial function like [tex]\( x^2 - 3x - 4 \)[/tex], the domain is all real numbers. There are no restrictions on [tex]\(x\)[/tex] for polynomial functions.
So, the domain of [tex]\((f + g)(x)\)[/tex] is:
[tex]\[ \text{Domain} = \text{All real numbers} \][/tex]
3. Determine the range of [tex]\((f + g)(x)\)[/tex]:
The range of a function is the set of all possible output values (y-values).
For a quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex], if the leading coefficient [tex]\( a \)[/tex] is positive, the parabola opens upwards and the range is [tex]\([k, \infty)\)[/tex], where [tex]\( k \)[/tex] is the minimum value of the function. Conversely, if [tex]\( a \)[/tex] is negative, the parabola opens downwards, but in this case, since the coefficient of [tex]\( x^2 \)[/tex] in [tex]\((f + g)(x) = x^2 - 3x - 4\)[/tex] is positive, the parabola opens upwards.
Since the quadratic term [tex]\( x^2 \)[/tex] dominates as [tex]\( x \)[/tex] goes to [tex]\( \pm \infty \)[/tex], the function [tex]\( (f + g)(x) = x^2 - 3x - 4 \)[/tex] will go to [tex]\( \infty \)[/tex] as [tex]\( x \)[/tex] goes to both [tex]\( \infty \)[/tex] and [tex]\( -\infty \)[/tex].
Therefore, the range covers all real numbers:
[tex]\[ \text{Range} = \text{All real numbers} \][/tex]
4. Some Example Calculations of [tex]\((f + g)(x)\)[/tex]:
Let's calculate [tex]\((f + g)(x)\)[/tex] for some example values of [tex]\( x \)[/tex]:
For [tex]\( x = 0 \)[/tex]:
[tex]\[ (f + g)(0) = 0^2 - 3(0) - 4 = -4 \][/tex]
For [tex]\( x = 1 \)[/tex]:
[tex]\[ (f + g)(1) = 1^2 - 3(1) - 4 = 1 - 3 - 4 = -6 \][/tex]
For [tex]\( x = 2 \)[/tex]:
[tex]\[ (f + g)(2) = 2^2 - 3(2) - 4 = 4 - 6 - 4 = -6 \][/tex]
Thus, for [tex]\( x \in \{0, 1, 2\} \)[/tex], the corresponding function values are:
[tex]\[ (f + g)(0) = -4 \][/tex]
[tex]\[ (f + g)(1) = -6 \][/tex]
[tex]\[ (f + g)(2) = -6 \][/tex]
So, the results we have for the function values are:
[tex]\[ x \in \{0, 1, 2\} \][/tex]
[tex]\[ (f + g)(x) \in \{-4, -6, -6\} \][/tex]
This brings us to the full solution.
Summary:
- The function [tex]\( f(x) + g(x) = x^2 - 3x - 4 \)[/tex]
- The domain is all real numbers.
- The range is all real numbers.
- Example calculations: [tex]\((f + g)(0) = -4\)[/tex], [tex]\((f + g)(1) = -6\)[/tex], [tex]\((f + g)(2) = -6\)[/tex].
