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Chapter 2: Problem 48

Solve the inequality. Then graph the solution set. $$\frac{1}{x-3} \leq \frac{9}{4 x+3}$$

Short Answer

Expert verified
The solution to the inequality is \(x \geq 6\).

Step by step solution

01

Get rid of denominators

To make the equation easier to handle, we'll want to get rid of the denominators. Multiply every term by \(x-3\) and \(4x+3\), leading to \(4x+3 \leq 9*(x-3)\)
02

Simplify the inequality

Expand the right hand side and combine like terms, resulting in \(4x+3 \leq 9x-27\). Then move all x terms to one side and constant terms to the other side to get \(5x \geq 30\). Thus, we obtain \(x \geq 6\)
03

Determine the critical points and test intervals

Next, we find the critical points by setting the denominators equal to zero: \(x-3=0\) gives \(x=3\) and \(4x+3=0\) gives \(x=-3/4\). For these points, the inequality is undefined. These points segment the number line into four parts: \(-\infty\) to \(-3/4\), \(-3/4\) to \(3\), \(3\) to \(6\), \(6\) to \(\infty\). Now, select a test point from each interval and substitute it into the original inequality to check the sign.
04

Graph the solution

From the sign check, we find that the intervals \(6\) to \(\infty\) make the original inequality true. Therefore, these are the intervals of the solution. So, the solution graph will have a filled circle at \(x = 6\) and will extend to the right towards \(\infty\).

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Most popular questions from this chapter

A cubic polynomial function \(f\) has real zeros \(-2, \frac{1}{2},\) and \(3,\) and its leading coefficient is negative. Write an equation for \(f\) and sketch its graph. How many different polynomial functions are possible for \(f ?\)

Use a graphing utility to graph the quadratic function. Identify the vertex, axis of symmetry, and \(x\) -intercept(s). Then check your results algebraically by writing the quadratic function in standard form. $$g(x)=\frac{1}{2}\left(x^{2}+4 x-2\right)$$

Match the cubic function with the numbers of rational and irrational zeros. (a) Rational zeros: \(0 ;\) irrational zeros: 1 (b) Rational zeros: \(3 ;\) irrational zeros: 0 (c) Rational zeros: \(1 ;\) irrational zeros: 2 (d) Rational zeros: \(1 ;\) irrational zeros: 0 $$f(x)=x^{3}-2$$

Sketch the graph of a fifth-degree polynomial function whose leading coefficient is positive and that has a zero at \(x=3\) of multiplicity 2.

Find the rational zeros of the polynomial function. $$f(x)=x^{3}-\frac{3}{2} x^{2}-\frac{23}{2} x+6=\frac{1}{2}\left(2 x^{3}-3 x^{2}-23 x+12\right)$$
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