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Chapter 4: Problem 15

For the function $$ h(x)=\frac{7 x}{(x-1)(x+5)} $$ solve each of the following. $$h(x) < 0$$

Short Answer

Expert verified
For \(0 < x < 1\), \(h(x) < 0\).

Step by step solution

01

- Find the roots of the numerator

Set the numerator equal to zero: \(7x = 0\). Solving this gives \(x = 0\). This is one of the critical points where the fraction's value is zero.
02

- Find the roots of the denominator

Set each factor of the denominator equal to zero: \((x-1) = 0\) and \((x+5) = 0\). Solving these gives the critical points \(x = 1\) and \(x = -5\). These values indicate where the function might have vertical asymptotes or undefined behavior.
03

- Determine the intervals

Using the critical points \(x = -5\), \(x = 0\) and \(x = 1\), divide the number line into the following intervals: \((-\infty, -5)\), \((-5, 0)\), \((0, 1)\), and \((1, \infty)\).
04

- Test the sign of \(h(x)\) in each interval

Pick a test point from each interval and substitute it into \(h(x) = \frac{7x}{(x-1)(x+5)}\) to determine the sign of the function in each interval:- For \((-\infty, -5)\), choose \(x = -6\): \(h(-6) = \frac{7(-6)}{((-6)-1)((-6)+5)} = \frac{-42}{-7} = 6\) (Positive).- For \((-5, 0)\), choose \(x = -1\): \(h(-1) = \frac{7(-1)}{((-1)-1)((-1)+5)} = \frac{-7}{-4} = 1.75\) (Positive).- For \((0, 1)\), choose \(x = 0.5\): \(h(0.5) = \frac{7(0.5)}{((0.5)-1)((0.5)+5)} = \frac{3.5}{-2.25} = -1.555\) (Negative).- For \((1, \infty)\), choose \(x = 2\): \(h(2) = \frac{7(2)}{((2)-1)((2)+5)} = \frac{14}{7} = 2\) (Positive).
05

- Identify where \(h(x) < 0\)

From Step 4, \(h(x)\) is negative in the interval \((0, 1)\). Therefore, \(h(x) < 0\) for \(0 < x < 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

critical points
Sign analysis is about determining whether the function is positive or negative in each interval. This is usually done by substituting test points from each interval into the function.

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Most popular questions from this chapter

What does Descartes' rule of signs tell you about the number of positive real zeros and the number of negative real zeros of the function? $$H(t)=5 t^{12}-7 t^{4}+3 t^{2}+t+1$$

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Determine the degree and the leading term of the polynomial function. $$f(x)=\left(x^{5}-1\right)^{2}\left(x^{2}+2\right)^{3}$$

Find a rational function that satisfies the given conditions. Answers may vary, but try to give the simplest answer possible. Vertical asymptotes \(x=-4, x=5\)
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