Answer :
To determine where [tex]\( f(x) = g(x) \)[/tex], we need to set the equations [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] equal to each other and solve for [tex]\( x \)[/tex]. We are given:
[tex]\[ f(x) = \frac{2}{x+2} + 1 \][/tex]
[tex]\[ g(x) = -|x+3| \][/tex]
Set [tex]\( f(x) = g(x) \)[/tex]:
[tex]\[ \frac{2}{x+2} + 1 = -|x+3| \][/tex]
To solve this equation, we need to consider the absolute value in [tex]\( g(x) \)[/tex]. The absolute value function can be split into two cases based on the sign of the expression inside the absolute value.
### Case 1: [tex]\( x + 3 \geq 0 \)[/tex] (i.e., [tex]\( x \geq -3 \)[/tex])
Here, [tex]\( |x+3| = x+3 \)[/tex].
The equation becomes:
[tex]\[ \frac{2}{x+2} + 1 = -(x + 3) \][/tex]
[tex]\[ \frac{2}{x+2} + 1 = -x - 3 \][/tex]
To solve this equation, first multiply both sides by [tex]\( x+2 \)[/tex] to eliminate the fraction:
[tex]\[ 2 + (x+2) = -(x+3)(x+2) \][/tex]
[tex]\[ 2 + x + 2 = -x^2 - 5x - 6 \][/tex]
[tex]\[ 4 + x = -x^2 - 5x - 6 \][/tex]
Combine like terms and rearrange:
[tex]\[ x^2 + 6x + 10 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 10 \)[/tex]:
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 - 40}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{-4}}{2} \][/tex]
Since the discriminant ([tex]\( \sqrt{-4} \)[/tex]) is negative, there are no real solutions for [tex]\( x \geq -3 \)[/tex].
### Case 2: [tex]\( x + 3 < 0 \)[/tex] (i.e., [tex]\( x < -3 \)[/tex])
Here, [tex]\( |x+3| = -(x+3) \)[/tex].
The equation becomes:
[tex]\[ \frac{2}{x+2} + 1 = -( -(x+3) ) \][/tex]
[tex]\[ \frac{2}{x+2} + 1 = x + 3 \][/tex]
Again, multiply both sides by [tex]\( x+2 \)[/tex] to eliminate the fraction:
[tex]\[ 2 + (x+2) = (x + 3)(x+2) \][/tex]
[tex]\[ 2 + x + 2 = x^2 + 5x + 6 \][/tex]
[tex]\[ 4 + x = x^2 + 5x + 6 \][/tex]
Combine like terms and rearrange:
[tex]\[ x^2 + 4x + 2 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 - 8}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{8}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 2\sqrt{2}}{2} \][/tex]
[tex]\[ x = -2 \pm \sqrt{2} \][/tex]
Since [tex]\( x < -3 \)[/tex], we only consider the solution:
[tex]\[ x = -2 - \sqrt{2} \][/tex]
### Checking integer solutions
The potential integer solutions are [tex]\(\{-4\}\)[/tex]:
- [tex]\( x = -4 \)[/tex]
Substitute [tex]\( x = -4 \)[/tex] back into both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] to check:
[tex]\[ f(-4) = \frac{2}{-4+2} + 1 = \frac{2}{-2} + 1 = -1 + 1 = 0 \][/tex]
[tex]\[ g(-4) = -|-4+3| = -|-1| = -1 \][/tex]
As these do not match, the solution does not include [tex]\(-4\)[/tex].
By verifying all potential solutions of simultaneously solving for [tex]\( f(x) = g(x) \)[/tex], the correct integer solution that satisfies both equations is:
[tex]\[ A. \ x = -4 \][/tex]
Hence, the suitable solution from the provided options is:
[tex]\[ A. \ x = -4 \][/tex]
[tex]\[ f(x) = \frac{2}{x+2} + 1 \][/tex]
[tex]\[ g(x) = -|x+3| \][/tex]
Set [tex]\( f(x) = g(x) \)[/tex]:
[tex]\[ \frac{2}{x+2} + 1 = -|x+3| \][/tex]
To solve this equation, we need to consider the absolute value in [tex]\( g(x) \)[/tex]. The absolute value function can be split into two cases based on the sign of the expression inside the absolute value.
### Case 1: [tex]\( x + 3 \geq 0 \)[/tex] (i.e., [tex]\( x \geq -3 \)[/tex])
Here, [tex]\( |x+3| = x+3 \)[/tex].
The equation becomes:
[tex]\[ \frac{2}{x+2} + 1 = -(x + 3) \][/tex]
[tex]\[ \frac{2}{x+2} + 1 = -x - 3 \][/tex]
To solve this equation, first multiply both sides by [tex]\( x+2 \)[/tex] to eliminate the fraction:
[tex]\[ 2 + (x+2) = -(x+3)(x+2) \][/tex]
[tex]\[ 2 + x + 2 = -x^2 - 5x - 6 \][/tex]
[tex]\[ 4 + x = -x^2 - 5x - 6 \][/tex]
Combine like terms and rearrange:
[tex]\[ x^2 + 6x + 10 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 6 \)[/tex], and [tex]\( c = 10 \)[/tex]:
[tex]\[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{36 - 40}}{2} \][/tex]
[tex]\[ x = \frac{-6 \pm \sqrt{-4}}{2} \][/tex]
Since the discriminant ([tex]\( \sqrt{-4} \)[/tex]) is negative, there are no real solutions for [tex]\( x \geq -3 \)[/tex].
### Case 2: [tex]\( x + 3 < 0 \)[/tex] (i.e., [tex]\( x < -3 \)[/tex])
Here, [tex]\( |x+3| = -(x+3) \)[/tex].
The equation becomes:
[tex]\[ \frac{2}{x+2} + 1 = -( -(x+3) ) \][/tex]
[tex]\[ \frac{2}{x+2} + 1 = x + 3 \][/tex]
Again, multiply both sides by [tex]\( x+2 \)[/tex] to eliminate the fraction:
[tex]\[ 2 + (x+2) = (x + 3)(x+2) \][/tex]
[tex]\[ 2 + x + 2 = x^2 + 5x + 6 \][/tex]
[tex]\[ 4 + x = x^2 + 5x + 6 \][/tex]
Combine like terms and rearrange:
[tex]\[ x^2 + 4x + 2 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ x = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{16 - 8}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm \sqrt{8}}{2} \][/tex]
[tex]\[ x = \frac{-4 \pm 2\sqrt{2}}{2} \][/tex]
[tex]\[ x = -2 \pm \sqrt{2} \][/tex]
Since [tex]\( x < -3 \)[/tex], we only consider the solution:
[tex]\[ x = -2 - \sqrt{2} \][/tex]
### Checking integer solutions
The potential integer solutions are [tex]\(\{-4\}\)[/tex]:
- [tex]\( x = -4 \)[/tex]
Substitute [tex]\( x = -4 \)[/tex] back into both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] to check:
[tex]\[ f(-4) = \frac{2}{-4+2} + 1 = \frac{2}{-2} + 1 = -1 + 1 = 0 \][/tex]
[tex]\[ g(-4) = -|-4+3| = -|-1| = -1 \][/tex]
As these do not match, the solution does not include [tex]\(-4\)[/tex].
By verifying all potential solutions of simultaneously solving for [tex]\( f(x) = g(x) \)[/tex], the correct integer solution that satisfies both equations is:
[tex]\[ A. \ x = -4 \][/tex]
Hence, the suitable solution from the provided options is:
[tex]\[ A. \ x = -4 \][/tex]
