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Given f of x is equal to 10 divided by the quantity x squared minus 7x minus 30, which of the following is true?

f(x) is negative for all x < –3
f(x) is negative for all x > –3
f(x) is positive for all x < 10
f(x) is positive for all x > 10

1 answer

To analyze the function \( f(x) = \frac{10}{x^2 - 7x - 30} \), we will first need to factor the denominator, \( x^2 - 7x - 30 \).

The quadratic can be factored as follows:

\[ x^2 - 7x - 30 = (x - 10)(x + 3) \]

Thus, we have:

\[ f(x) = \frac{10}{(x - 10)(x + 3)} \]

Now, we can determine the intervals where \( f(x) \) is positive or negative by examining the sign of the denominator, \( (x - 10)(x + 3) \).

Step 1: Find the zeros of the denominator

The denominator is zero when:

\[ (x - 10)(x + 3) = 0 \implies x = 10 \text{ or } x = -3 \]

Step 2: Analyze the sign of \( f(x) \)

The critical points divide the real line into three intervals: \( (-\infty, -3) \), \( (-3, 10) \), and \( (10, \infty) \).

  1. Interval \( (-\infty, -3) \):

    • Choose a test point, e.g., \( x = -4 \): \[ f(-4) = \frac{10}{(-4 - 10)(-4 + 3)} = \frac{10}{(-14)(-1)} = \frac{10}{14} > 0 \] So, \( f(x) > 0 \) in this interval.

  2. Interval \( (-3, 10) \):

    • Choose a test point, e.g., \( x = 0 \): \[ f(0) = \frac{10}{(0 - 10)(0 + 3)} = \frac{10}{(-10)(3)} = \frac{10}{-30} < 0 \] So, \( f(x) < 0 \) in this interval.

  3. Interval \( (10, \infty) \):

    • Choose a test point, e.g., \( x = 11 \): \[ f(11) = \frac{10}{(11 - 10)(11 + 3)} = \frac{10}{(1)(14)} = \frac{10}{14} > 0 \] So, \( f(x) > 0 \) in this interval.

Step 3: Summary of intervals

  • \( f(x) > 0 \) for \( x < -3 \)
  • \( f(x) < 0 \) for \( -3 < x < 10 \)
  • \( f(x) > 0 \) for \( x > 10 \)

Final answer verification

Based on the analysis, we can confirm:

  • \( f(x) \) is negative for all \( x > -3 \) and \( x < 10 \) (not for all \( x < -3 \)).
  • Thus, the only statements we can derive from the choices are:
    • \( f(x) \) is negative for \( -3 < x < 10 \), which doesn't match any given answer.
    • \( f(x) \) is positive for \( x < -3 \) and \( x > 10 \), but not all of those intervals match.

Therefore, none of the provided choices are correct in their entirety as stated.

However, based on available options, we could endorse:

  • None is strictly true. But if we were to choose the closest "-3" product, we could tentatively say \( f(x) \) is positive for all \( x < 10 \) is not wholly incorrect in terms of its domain break at -3 but remains false in the wider continuous evaluation.

Thus, the true evaluations of positivity hold between defined ranges continuously by the quadratic analysis alone.