Chapter 7: Q. 501. (page 742)

Given the function, $R (x) = \frac{2}{2 x^{2} + x - 15}$ , find the values of $x$ that make the function less than or equal to $0$ .

Short Answer

Expert verified

The solution set of the given inequality will be $x \in (- 3, \frac{5}{2})$ .

Step by step solution

Step 1. Given information.

The given expression is $R (x) = \frac{2}{2 x^{2} + x - 15}$ .

Step 2. Solve Rx≤0.

$\frac{2}{2 x^{2} + x - 15} \leq 0$

This implies that $2 x^{2} + x - 15$ must be $< 0$ .

Now set the denominator equal to zero to get the critical points.

$2 x^{2} + x - 15 = 0 2 x^{2} + 6 x - 5 x - 15 = 0 2 x (x + 3) - 5 (x + 3) = 0 (x + 3) (2 x - 5) = 0 \Rightarrow x = - 3 o r x = \frac{5}{2}$

Step 3. Now text the points between the critical points as well as out the critical points.

At $x = - 5$ .

$R (- 5) = \frac{2}{2 {(- 5)}^{2} + (- 5) - 15} R (- 5) = \frac{1}{15}$

At $x = 0$ .

$R (0) = \frac{2}{2 {(0)}^{2} + (0) - 15} R (0) = - \frac{2}{15}$

At $x = 3$ .

$R (3) = \frac{2}{2 {(3)}^{2} + 3 - 15} R (3) = \frac{1}{3}$

Step 4. Simplified answer.

Hence, the solution set of the given inequality will be $x \in (- 3, \frac{5}{2})$ .

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Given the function, $R (x) = \frac{2}{2 x^{2} + x - 15}$ , find the values of $x$ that make the function less than or equal to $0$ .

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Solve Rx≤0.

Step 3. Now text the points between the critical points as well as out the critical points.

Step 4. Simplified answer.

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Given the function, Rx=22x2+x-15, find the values ofxthat make the function less than or equal to0.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Solve Rx≤0.

Step 3. Now text the points between the critical points as well as out the critical points.

Step 4. Simplified answer.

One App. One Place for Learning.

Most popular questions from this chapter

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Given the function, $R (x) = \frac{2}{2 x^{2} + x - 15}$ , find the values of $x$ that make the function less than or equal to $0$ .