Answer :
To solve the inequality [tex]\(\frac{5x + 1}{(x + 1)^2} \leq 1\)[/tex], let's go through a step-by-step solution:
1. Write the Inequality:
[tex]\[\frac{5x + 1}{(x + 1)^2} \leq 1.\][/tex]
2. Subtract 1 from Both Sides:
[tex]\[\frac{5x + 1}{(x + 1)^2} - 1 \leq 0.\][/tex]
3. Combine the Terms on the Left Side:
To combine the terms, we use a common denominator:
[tex]\[\frac{5x + 1}{(x + 1)^2} - \frac{(x + 1)^2}{(x + 1)^2} \leq 0.\][/tex]
This simplifies to:
[tex]\[\frac{5x + 1 - (x + 1)^2}{(x + 1)^2} \leq 0.\][/tex]
4. Simplify the Numerator:
Expand and simplify the numerator:
[tex]\[5x + 1 - (x^2 + 2x + 1) = 5x + 1 - x^2 - 2x - 1 = -x^2 + 3x.\][/tex]
So the inequality becomes:
[tex]\[\frac{-x^2 + 3x}{(x + 1)^2} \leq 0.\][/tex]
5. Factor the Numerator:
Factor [tex]\(-x^2 + 3x\)[/tex]:
[tex]\[-x^2 + 3x = x(3 - x).\][/tex]
The inequality now becomes:
[tex]\[\frac{x(3 - x)}{(x + 1)^2} \leq 0.\][/tex]
6. Determine the Critical Points:
[tex]\[x(3 - x) = 0 \quad \text{and} \quad (x + 1)^2 \neq 0.\][/tex]
The critical points are [tex]\(x = 0\)[/tex] and [tex]\(x = 3\)[/tex]. Note that [tex]\((x + 1)^2 = 0\)[/tex] gives [tex]\(x = -1\)[/tex], which is another critical point affecting the sign changes but does not make the numerator zero.
7. Test Intervals Around Critical Points:
Split the number line based on these points and test the inequality in each interval:
- For [tex]\(x < -1\)[/tex]
- For [tex]\(-1 < x < 0\)[/tex]
- For [tex]\(0 < x < 3\)[/tex]
- For [tex]\(x > 3\)[/tex]
8. Check Sign Changes:
Analyze the sign of the expression [tex]\(\frac{x(3 - x)}{(x + 1)^2}\)[/tex] in the intervals:
- For [tex]\(x < -1\)[/tex]
Both [tex]\(x\)[/tex] and [tex]\(3 - x\)[/tex] are negative, and [tex]\((x + 1)^2\)[/tex] is positive, making the expression positive.
- For [tex]\(-1 < x < 0\)[/tex]
[tex]\(x\)[/tex] is negative, [tex]\(3 - x\)[/tex] is positive, and [tex]\((x + 1)^2\)[/tex] is positive, making the expression negative.
- For [tex]\(0 < x < 3\)[/tex]
Both [tex]\(x\)[/tex] and [tex]\(3 - x\)[/tex] are positive, and [tex]\((x + 1)^2\)[/tex] is positive, making the expression positive.
- For [tex]\(x > 3\)[/tex]
[tex]\(x\)[/tex] and [tex]\(3 - x\)[/tex] are of opposite signs, making the expression negative.
9. Include Boundary Points:
- For [tex]\(x = -1\)[/tex], the denominator becomes zero, and the expression is undefined.
- For [tex]\(x = 0\)[/tex] and [tex]\(x = 3\)[/tex], the numerator is zero, making the entire fraction zero.
10. Combine the Intervals:
From the analysis, the expression [tex]\(\frac{x(3 - x)}{(x + 1)^2} \leq 0\)[/tex] is satisfied for:
- [tex]\(-\infty < x < -1\)[/tex] (open interval, expression positive)
- [tex]\(-1 < x \leq 0\)[/tex] (expression non-positive)
- [tex]\(3 \leq x < \infty\)[/tex] (expression non-positive)
Thus, the set of values of [tex]\(x\)[/tex] that satisfy the inequality is:
[tex]\[\boxed{(-\infty, -1) \cup (-1, 0] \cup [3, \infty)}.\][/tex]
1. Write the Inequality:
[tex]\[\frac{5x + 1}{(x + 1)^2} \leq 1.\][/tex]
2. Subtract 1 from Both Sides:
[tex]\[\frac{5x + 1}{(x + 1)^2} - 1 \leq 0.\][/tex]
3. Combine the Terms on the Left Side:
To combine the terms, we use a common denominator:
[tex]\[\frac{5x + 1}{(x + 1)^2} - \frac{(x + 1)^2}{(x + 1)^2} \leq 0.\][/tex]
This simplifies to:
[tex]\[\frac{5x + 1 - (x + 1)^2}{(x + 1)^2} \leq 0.\][/tex]
4. Simplify the Numerator:
Expand and simplify the numerator:
[tex]\[5x + 1 - (x^2 + 2x + 1) = 5x + 1 - x^2 - 2x - 1 = -x^2 + 3x.\][/tex]
So the inequality becomes:
[tex]\[\frac{-x^2 + 3x}{(x + 1)^2} \leq 0.\][/tex]
5. Factor the Numerator:
Factor [tex]\(-x^2 + 3x\)[/tex]:
[tex]\[-x^2 + 3x = x(3 - x).\][/tex]
The inequality now becomes:
[tex]\[\frac{x(3 - x)}{(x + 1)^2} \leq 0.\][/tex]
6. Determine the Critical Points:
[tex]\[x(3 - x) = 0 \quad \text{and} \quad (x + 1)^2 \neq 0.\][/tex]
The critical points are [tex]\(x = 0\)[/tex] and [tex]\(x = 3\)[/tex]. Note that [tex]\((x + 1)^2 = 0\)[/tex] gives [tex]\(x = -1\)[/tex], which is another critical point affecting the sign changes but does not make the numerator zero.
7. Test Intervals Around Critical Points:
Split the number line based on these points and test the inequality in each interval:
- For [tex]\(x < -1\)[/tex]
- For [tex]\(-1 < x < 0\)[/tex]
- For [tex]\(0 < x < 3\)[/tex]
- For [tex]\(x > 3\)[/tex]
8. Check Sign Changes:
Analyze the sign of the expression [tex]\(\frac{x(3 - x)}{(x + 1)^2}\)[/tex] in the intervals:
- For [tex]\(x < -1\)[/tex]
Both [tex]\(x\)[/tex] and [tex]\(3 - x\)[/tex] are negative, and [tex]\((x + 1)^2\)[/tex] is positive, making the expression positive.
- For [tex]\(-1 < x < 0\)[/tex]
[tex]\(x\)[/tex] is negative, [tex]\(3 - x\)[/tex] is positive, and [tex]\((x + 1)^2\)[/tex] is positive, making the expression negative.
- For [tex]\(0 < x < 3\)[/tex]
Both [tex]\(x\)[/tex] and [tex]\(3 - x\)[/tex] are positive, and [tex]\((x + 1)^2\)[/tex] is positive, making the expression positive.
- For [tex]\(x > 3\)[/tex]
[tex]\(x\)[/tex] and [tex]\(3 - x\)[/tex] are of opposite signs, making the expression negative.
9. Include Boundary Points:
- For [tex]\(x = -1\)[/tex], the denominator becomes zero, and the expression is undefined.
- For [tex]\(x = 0\)[/tex] and [tex]\(x = 3\)[/tex], the numerator is zero, making the entire fraction zero.
10. Combine the Intervals:
From the analysis, the expression [tex]\(\frac{x(3 - x)}{(x + 1)^2} \leq 0\)[/tex] is satisfied for:
- [tex]\(-\infty < x < -1\)[/tex] (open interval, expression positive)
- [tex]\(-1 < x \leq 0\)[/tex] (expression non-positive)
- [tex]\(3 \leq x < \infty\)[/tex] (expression non-positive)
Thus, the set of values of [tex]\(x\)[/tex] that satisfy the inequality is:
[tex]\[\boxed{(-\infty, -1) \cup (-1, 0] \cup [3, \infty)}.\][/tex]
