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Sagot :
To determine for which ranges [tex]\( f(x) \)[/tex] is positive or negative, we need to analyze the function [tex]\( f(x) = \frac{x+4}{x^2 - 3x - 28} \)[/tex]. Here’s a detailed step-by-step solution:
1. Find the zeroes of the numerator: We set the numerator [tex]\( x + 4 \)[/tex] equal to zero:
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
So, the zero of the numerator is [tex]\( x = -4 \)[/tex].
2. Find the zeroes of the denominator: We set the denominator [tex]\( x^2 - 3x - 28 \)[/tex] equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 3x - 28 = 0 \][/tex]
Solving this quadratic equation:
[tex]\[ x = \frac{3 \pm \sqrt{9 + 112}}{2} = \frac{3 \pm \sqrt{121}}{2} = \frac{3 \pm 11}{2} \][/tex]
This gives:
[tex]\[ x = \frac{3 + 11}{2} = 7 \quad \text{and} \quad x = \frac{3 - 11}{2} = -4 \][/tex]
Hence, the zeroes of the denominator are [tex]\( x = -4 \)[/tex] and [tex]\( x = 7 \)[/tex].
3. Identify the critical points and their intervals: The critical points where the function changes behavior are [tex]\( x = -4 \)[/tex] and [tex]\( x = 7 \)[/tex]. These points divide the real number line into the intervals:
[tex]\[ (-\infty, -4), \quad (-4, 7), \quad (7, \infty) \][/tex]
4. Test the sign of [tex]\( f(x) \)[/tex] in each interval: We choose test points in each interval and check the sign of [tex]\( f(x) \)[/tex].
- For the interval [tex]\((- \infty, -4) \)[/tex], we choose [tex]\( x = -10 \)[/tex]:
[tex]\[ f(-10) = \frac{-10 + 4}{(-10)^2 - 3(-10) - 28} = \frac{-6}{100 - 30 - 28} = \frac{-6}{42} < 0 \][/tex]
So, [tex]\( f(x) \)[/tex] is negative in [tex]\((- \infty, -4) \)[/tex].
- For the interval [tex]\( (-4, 7) \)[/tex], we choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{0 + 4}{0^2 - 3(0) - 28} = \frac{4}{-28} = -\frac{1}{7} < 0 \][/tex]
So, [tex]\( f(x) \)[/tex] is negative in [tex]\((-4, 7) \)[/tex].
- For the interval [tex]\( (7, \infty) \)[/tex], we choose [tex]\( x = 8 \)[/tex]:
[tex]\[ f(8) = \frac{8 + 4}{8^2 - 3(8) - 28} = \frac{12}{64 - 24 - 28} = \frac{12}{12} = 1 > 0 \][/tex]
So, [tex]\( f(x) \)[/tex] is positive in [tex]\( (7, \infty) \)[/tex].
Since [tex]\( f(-4) \)[/tex] is not defined because it makes the denominator zero, we exclude [tex]\( x = -4 \)[/tex].
5. Conclusions about the sign of [tex]\( f(x) \)[/tex]:
- [tex]\( f(x) \)[/tex] is negative for [tex]\( x < -4 \)[/tex] and [tex]\( -4 < x < 7 \)[/tex].
- [tex]\( f(x) \)[/tex] is positive for [tex]\( x > 7 \)[/tex].
From the intervals where we determined the sign of [tex]\( f(x) \)[/tex], we can conclude that:
- [tex]\( f(x) \)[/tex] is negative for [tex]\( x < 7 \)[/tex].
- [tex]\( f(x) \)[/tex] is not positive for all [tex]\( x > -4 \)[/tex], nor negative for all [tex]\( x > -4 \)[/tex].
Thus, the correct conclusion is:
[tex]\[ \text{f(x) is negative for all } x < 7.\][/tex]
1. Find the zeroes of the numerator: We set the numerator [tex]\( x + 4 \)[/tex] equal to zero:
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
So, the zero of the numerator is [tex]\( x = -4 \)[/tex].
2. Find the zeroes of the denominator: We set the denominator [tex]\( x^2 - 3x - 28 \)[/tex] equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 - 3x - 28 = 0 \][/tex]
Solving this quadratic equation:
[tex]\[ x = \frac{3 \pm \sqrt{9 + 112}}{2} = \frac{3 \pm \sqrt{121}}{2} = \frac{3 \pm 11}{2} \][/tex]
This gives:
[tex]\[ x = \frac{3 + 11}{2} = 7 \quad \text{and} \quad x = \frac{3 - 11}{2} = -4 \][/tex]
Hence, the zeroes of the denominator are [tex]\( x = -4 \)[/tex] and [tex]\( x = 7 \)[/tex].
3. Identify the critical points and their intervals: The critical points where the function changes behavior are [tex]\( x = -4 \)[/tex] and [tex]\( x = 7 \)[/tex]. These points divide the real number line into the intervals:
[tex]\[ (-\infty, -4), \quad (-4, 7), \quad (7, \infty) \][/tex]
4. Test the sign of [tex]\( f(x) \)[/tex] in each interval: We choose test points in each interval and check the sign of [tex]\( f(x) \)[/tex].
- For the interval [tex]\((- \infty, -4) \)[/tex], we choose [tex]\( x = -10 \)[/tex]:
[tex]\[ f(-10) = \frac{-10 + 4}{(-10)^2 - 3(-10) - 28} = \frac{-6}{100 - 30 - 28} = \frac{-6}{42} < 0 \][/tex]
So, [tex]\( f(x) \)[/tex] is negative in [tex]\((- \infty, -4) \)[/tex].
- For the interval [tex]\( (-4, 7) \)[/tex], we choose [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{0 + 4}{0^2 - 3(0) - 28} = \frac{4}{-28} = -\frac{1}{7} < 0 \][/tex]
So, [tex]\( f(x) \)[/tex] is negative in [tex]\((-4, 7) \)[/tex].
- For the interval [tex]\( (7, \infty) \)[/tex], we choose [tex]\( x = 8 \)[/tex]:
[tex]\[ f(8) = \frac{8 + 4}{8^2 - 3(8) - 28} = \frac{12}{64 - 24 - 28} = \frac{12}{12} = 1 > 0 \][/tex]
So, [tex]\( f(x) \)[/tex] is positive in [tex]\( (7, \infty) \)[/tex].
Since [tex]\( f(-4) \)[/tex] is not defined because it makes the denominator zero, we exclude [tex]\( x = -4 \)[/tex].
5. Conclusions about the sign of [tex]\( f(x) \)[/tex]:
- [tex]\( f(x) \)[/tex] is negative for [tex]\( x < -4 \)[/tex] and [tex]\( -4 < x < 7 \)[/tex].
- [tex]\( f(x) \)[/tex] is positive for [tex]\( x > 7 \)[/tex].
From the intervals where we determined the sign of [tex]\( f(x) \)[/tex], we can conclude that:
- [tex]\( f(x) \)[/tex] is negative for [tex]\( x < 7 \)[/tex].
- [tex]\( f(x) \)[/tex] is not positive for all [tex]\( x > -4 \)[/tex], nor negative for all [tex]\( x > -4 \)[/tex].
Thus, the correct conclusion is:
[tex]\[ \text{f(x) is negative for all } x < 7.\][/tex]
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