To solve for the values of [tex]\( x \)[/tex] where [tex]\( f(x) = g(x) \)[/tex], we need to solve the equation
[tex]\[ x^2 - 3x - 4 = |x + 3| - 2. \][/tex]
First, let's understand the function [tex]\( g(x) = |x + 3| - 2 \)[/tex]. The absolute value function [tex]\( |x + 3| \)[/tex] has two cases depending on the value of [tex]\( x \)[/tex]:
1. If [tex]\( x + 3 \geq 0 \)[/tex] (i.e., [tex]\( x \geq -3 \)[/tex]), then [tex]\( |x + 3| = x + 3 \)[/tex].
2. If [tex]\( x + 3 < 0 \)[/tex] (i.e., [tex]\( x < -3 \)[/tex]), then [tex]\( |x + 3| = -(x + 3) = -x - 3 \)[/tex].
Let's look at each case separately:
### Case 1: [tex]\( x \geq -3 \)[/tex]
Here, [tex]\( |x + 3| = x + 3 \)[/tex]. Substituting this into [tex]\( g(x) \)[/tex], we get:
[tex]\[ g(x) = (x + 3) - 2 = x + 1. \][/tex]
So, we need to solve the equation:
[tex]\[ x^2 - 3x - 4 = x + 1. \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ x^2 - 3x - 4 - x - 1 = 0 \][/tex]
[tex]\[ x^2 - 4x - 5 = 0. \][/tex]
Next, solve this quadratic equation by factoring:
[tex]\[ x^2 - 4x - 5 = (x - 5)(x + 1) = 0. \][/tex]
Thus, the solutions are:
[tex]\[ x = 5 \quad \text{or} \quad x = -1. \][/tex]
We need to check if these solutions lie within the range [tex]\( x \geq -3 \)[/tex]:
- [tex]\( x = 5 \)[/tex] is valid.
- [tex]\( x = -1 \)[/tex] is valid.
### Case 2: [tex]\( x < -3 \)[/tex]
Here, [tex]\( |x + 3| = -x - 3 \)[/tex]. Substituting this into [tex]\( g(x) \)[/tex], we get:
[tex]\[ g(x) = (-x - 3) - 2 = -x - 5. \][/tex]
So, we need to solve the equation:
[tex]\[ x^2 - 3x - 4 = -x - 5. \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ x^2 - 3x - 4 + x + 5 = 0 \][/tex]
[tex]\[ x^2 - 2x + 1 = 0. \][/tex]
Next, solve this quadratic equation by factoring:
[tex]\[ x^2 - 2x + 1 = (x - 1)^2 = 0. \][/tex]
Thus, the solution is:
[tex]\[ x = 1. \][/tex]
We need to check if this solution lies within the range [tex]\( x < -3 \)[/tex]:
- [tex]\( x = 1 \)[/tex] is not valid in this case.
Hence, the only valid solutions are [tex]\( x = 5 \)[/tex] and [tex]\( x = -1 \)[/tex].
From the given options:
A. -4
B. -3
C. -1
D. 4
The valid solution present in the options is:
[tex]\[ \boxed{-1} \][/tex]
