Answer :
Here's how we can solve the inequality [tex]\(f(x) \cdot g(x) \leq 0\)[/tex] for the given functions [tex]\(f(x)=\frac{2}{x+1}+3\)[/tex] and [tex]\(g(x)=2x+5\)[/tex]:
### Step 1: Find the zeros of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]
First, solve for [tex]\(x\)[/tex] where each function is equal to zero:
f(x) = 0:
[tex]\[ \frac{2}{x + 1} + 3 = 0 \][/tex]
Subtract 3 from both sides:
[tex]\[ \frac{2}{x + 1} = -3 \][/tex]
Multiply both sides by [tex]\(x + 1\)[/tex]:
[tex]\[ 2 = -3(x + 1) \][/tex]
Simplify and solve for [tex]\(x\)[/tex]:
[tex]\[ 2 = -3x - 3 \\ 5 = -3x \\ x = -\frac{5}{3} \][/tex]
So, [tex]\(f(x) = 0\)[/tex] at [tex]\(x = -\frac{5}{3}\)[/tex].
g(x) = 0:
[tex]\[ 2x + 5 = 0 \][/tex]
Subtract 5 from both sides:
[tex]\[ 2x = -5 \][/tex]
Divide by 2:
[tex]\[ x = -\frac{5}{2} \][/tex]
So, [tex]\(g(x) = 0\)[/tex] at [tex]\(x = -\frac{5}{2}\)[/tex].
### Step 2: Identify additional critical points
We also need to find points where [tex]\( f(x) \)[/tex] or [tex]\( g(x) \)[/tex] may change signs. For [tex]\(f(x)\)[/tex], this includes its vertical asymptote:
The vertical asymptote of [tex]\( f(x) \)[/tex] is found by setting the denominator equal to zero:
[tex]\[ x + 1 = 0 \\ x = -1 \][/tex]
So the vertical asymptote is at [tex]\(x = -1\)[/tex].
### Step 3: Organize critical points and test intervals
The critical points we've identified are [tex]\(x = -\frac{5}{2}\)[/tex], [tex]\(x = -\frac{5}{3}\)[/tex], and [tex]\(x = -1\)[/tex].
Sort these critical points:
[tex]\[ x = -\frac{5}{2}, -\frac{5}{3}, -1 \][/tex]
These divide the number line into several intervals that we need to test:
1. [tex]\( x < -\frac{5}{2} \)[/tex]
2. [tex]\( -\frac{5}{2} < x < -\frac{5}{3} \)[/tex]
3. [tex]\( -\frac{5}{3} < x < -1 \)[/tex]
4. [tex]\( x > -1 \)[/tex]
### Step 4: Test intervals to determine where [tex]\(f(x) \cdot g(x) \leq 0\)[/tex]
Let's analyze the intervals:
- Interval [tex]\((- \infty, -\frac{5}{2})\)[/tex]:
Test point [tex]\(x = -3\)[/tex]:
[tex]\[ f(-3) = \frac{2}{-3 + 1} + 3 = \frac{2}{-2} + 3 = -1 + 3 = 2 \\ g(-3) = 2(-3) + 5 = -6 + 5 = -1 \][/tex]
[tex]\(f(-3) \cdot g(-3) = 2 \cdot (-1) = -2 \leq 0\)[/tex]
This interval satisfies the inequality.
- Interval [tex]\((- \frac{5}{2}, - \frac{5}{3})\)[/tex]:
Test point [tex]\(x = -2\)[/tex]:
[tex]\[ f(-2) = \frac{2}{-2 + 1} + 3 = \frac{2}{-1} + 3 = -2 + 3 = 1 \\ g(-2) = 2(-2) + 5 = -4 + 5 = 1 \][/tex]
[tex]\(f(-2) \cdot g(-2) = 1 \cdot 1 = 1 > 0\)[/tex]
This interval does not satisfy the inequality.
- Interval [tex]\((- \frac{5}{3}, -1)\)[/tex]:
Test point [tex]\(x = -\frac{4}{3}\)[/tex]:
[tex]\[ f\left(-\frac{4}{3}\right) = \frac{2}{-\frac{4}{3} + 1} + 3 = \frac{2}{-\frac{1}{3}} + 3 = -6 + 3 = -3 \\ g\left(-\frac{4}{3}\right) = 2\left(-\frac{4}{3}\right) + 5 = -\frac{8}{3} + 5 = -\frac{8}{3} + \frac{15}{3} = \frac{7}{3} \][/tex]
[tex]\(f\left(-\frac{4}{3}\right) \cdot g\left(-\frac{4}{3}\right) = -3 \cdot \frac{7}{3} = -7 \leq 0\)[/tex]
This interval satisfies the inequality.
- Interval [tex]\((-1, \infty)\)[/tex]:
Test point [tex]\(x = 0\)[/tex]:
[tex]\[ f(0) = \frac{2}{0 + 1} + 3 = 2 + 3 = 5 \\ g(0) = 2(0) + 5 = 5 \][/tex]
[tex]\(f(0) \cdot g(0) = 5 \cdot 5 = 25 > 0\)[/tex]
This interval does not satisfy the inequality.
### Step 5: Combine the results
The inequality [tex]\( f(x) \cdot g(x) \leq 0 \)[/tex] is satisfied in the intervals:
[tex]\[ (-\infty, -\frac{5}{2}) \cup (-\frac{5}{3}, -1) \][/tex]
So the solution to the inequality [tex]\(f(x) \cdot g(x) \leq 0\)[/tex] is [tex]\( x \in (-\infty, -\frac{5}{2}) \cup (-\frac{5}{3}, -1) \)[/tex].
### Step 1: Find the zeros of [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex]
First, solve for [tex]\(x\)[/tex] where each function is equal to zero:
f(x) = 0:
[tex]\[ \frac{2}{x + 1} + 3 = 0 \][/tex]
Subtract 3 from both sides:
[tex]\[ \frac{2}{x + 1} = -3 \][/tex]
Multiply both sides by [tex]\(x + 1\)[/tex]:
[tex]\[ 2 = -3(x + 1) \][/tex]
Simplify and solve for [tex]\(x\)[/tex]:
[tex]\[ 2 = -3x - 3 \\ 5 = -3x \\ x = -\frac{5}{3} \][/tex]
So, [tex]\(f(x) = 0\)[/tex] at [tex]\(x = -\frac{5}{3}\)[/tex].
g(x) = 0:
[tex]\[ 2x + 5 = 0 \][/tex]
Subtract 5 from both sides:
[tex]\[ 2x = -5 \][/tex]
Divide by 2:
[tex]\[ x = -\frac{5}{2} \][/tex]
So, [tex]\(g(x) = 0\)[/tex] at [tex]\(x = -\frac{5}{2}\)[/tex].
### Step 2: Identify additional critical points
We also need to find points where [tex]\( f(x) \)[/tex] or [tex]\( g(x) \)[/tex] may change signs. For [tex]\(f(x)\)[/tex], this includes its vertical asymptote:
The vertical asymptote of [tex]\( f(x) \)[/tex] is found by setting the denominator equal to zero:
[tex]\[ x + 1 = 0 \\ x = -1 \][/tex]
So the vertical asymptote is at [tex]\(x = -1\)[/tex].
### Step 3: Organize critical points and test intervals
The critical points we've identified are [tex]\(x = -\frac{5}{2}\)[/tex], [tex]\(x = -\frac{5}{3}\)[/tex], and [tex]\(x = -1\)[/tex].
Sort these critical points:
[tex]\[ x = -\frac{5}{2}, -\frac{5}{3}, -1 \][/tex]
These divide the number line into several intervals that we need to test:
1. [tex]\( x < -\frac{5}{2} \)[/tex]
2. [tex]\( -\frac{5}{2} < x < -\frac{5}{3} \)[/tex]
3. [tex]\( -\frac{5}{3} < x < -1 \)[/tex]
4. [tex]\( x > -1 \)[/tex]
### Step 4: Test intervals to determine where [tex]\(f(x) \cdot g(x) \leq 0\)[/tex]
Let's analyze the intervals:
- Interval [tex]\((- \infty, -\frac{5}{2})\)[/tex]:
Test point [tex]\(x = -3\)[/tex]:
[tex]\[ f(-3) = \frac{2}{-3 + 1} + 3 = \frac{2}{-2} + 3 = -1 + 3 = 2 \\ g(-3) = 2(-3) + 5 = -6 + 5 = -1 \][/tex]
[tex]\(f(-3) \cdot g(-3) = 2 \cdot (-1) = -2 \leq 0\)[/tex]
This interval satisfies the inequality.
- Interval [tex]\((- \frac{5}{2}, - \frac{5}{3})\)[/tex]:
Test point [tex]\(x = -2\)[/tex]:
[tex]\[ f(-2) = \frac{2}{-2 + 1} + 3 = \frac{2}{-1} + 3 = -2 + 3 = 1 \\ g(-2) = 2(-2) + 5 = -4 + 5 = 1 \][/tex]
[tex]\(f(-2) \cdot g(-2) = 1 \cdot 1 = 1 > 0\)[/tex]
This interval does not satisfy the inequality.
- Interval [tex]\((- \frac{5}{3}, -1)\)[/tex]:
Test point [tex]\(x = -\frac{4}{3}\)[/tex]:
[tex]\[ f\left(-\frac{4}{3}\right) = \frac{2}{-\frac{4}{3} + 1} + 3 = \frac{2}{-\frac{1}{3}} + 3 = -6 + 3 = -3 \\ g\left(-\frac{4}{3}\right) = 2\left(-\frac{4}{3}\right) + 5 = -\frac{8}{3} + 5 = -\frac{8}{3} + \frac{15}{3} = \frac{7}{3} \][/tex]
[tex]\(f\left(-\frac{4}{3}\right) \cdot g\left(-\frac{4}{3}\right) = -3 \cdot \frac{7}{3} = -7 \leq 0\)[/tex]
This interval satisfies the inequality.
- Interval [tex]\((-1, \infty)\)[/tex]:
Test point [tex]\(x = 0\)[/tex]:
[tex]\[ f(0) = \frac{2}{0 + 1} + 3 = 2 + 3 = 5 \\ g(0) = 2(0) + 5 = 5 \][/tex]
[tex]\(f(0) \cdot g(0) = 5 \cdot 5 = 25 > 0\)[/tex]
This interval does not satisfy the inequality.
### Step 5: Combine the results
The inequality [tex]\( f(x) \cdot g(x) \leq 0 \)[/tex] is satisfied in the intervals:
[tex]\[ (-\infty, -\frac{5}{2}) \cup (-\frac{5}{3}, -1) \][/tex]
So the solution to the inequality [tex]\(f(x) \cdot g(x) \leq 0\)[/tex] is [tex]\( x \in (-\infty, -\frac{5}{2}) \cup (-\frac{5}{3}, -1) \)[/tex].
