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Chapter 0: Problem 22

Find the domain of the function. $$ f(x)=\frac{\sqrt{x-1}}{x^{2}-x-6} $$

Short Answer

Expert verified
The domain of the function \(f(x)=\frac{\sqrt{x-1}}{x^{2}-x-6}\) is \(x \in (-\infty, -2) \cup (-2, 1) \cup (1, 3) \cup (3, \infty)\).

Step by step solution

01

Determine the range of the expression inside the square root

The expression inside the square root is \(x-1\). In order for the square root to be defined, the expression inside must be greater than or equal to 0. So we'll solve the inequality: $$x - 1 \geq 0$$
02

Solve the inequality

To solve for the inequality, add 1 to both sides: $$x \geq 1$$ Now we know that \(x\) must be greater or equal to 1 for the square root to be defined.
03

Identify the restriction on the denominator

The denominator of the function is \(x^2 - x - 6\). In order for the function to be defined, the denominator cannot be equal to 0. So, we need to find the zeroes of the denominator, as they're not part of the domain. To find the zeroes, set the denominator equal to 0: $$x^2 - x - 6 = 0$$
04

Solve the quadratic equation

Factor the quadratic equation by searching for two numbers that multiply to -6 and add to -1. In this case, those two numbers are -3 and 2. Therefore, the factored form of the quadratic is: $$(x - 3)(x + 2) = 0$$ Solve for \(x\): $$x = 3, -2$$ These are the two values that make the denominator of the function equal to 0 and cannot be part of the domain.
05

Combine the two restrictions

Combine the two restrictions to find the domain of the function. Recall that \(x\) must be greater or equal to 1 and cannot be equal to 3 or -2. So the domain is: $$x \in (-\infty, -2) \cup (-2, 1) \cup (1, 3) \cup (3, \infty)$$

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Most popular questions from this chapter

Sketch the graph of the first function by plotting points if necessary. Then use transformation(s) to obtain the graph of the second function. \(y=x^{2}, \quad y=\left|x^{2}-1\right|\)

Find the zero(s) of the function f to five decimal places. $$ f(x)=2 x^{4}-4 x^{2}+1 $$

Show that \(f\) and \(g\) are inverses of each other by verifying that \(f[g(x)]=x\) and \(g[f(x)]=x\). $$ \begin{aligned} &f(x)=4(x+1)^{2 / 3}, \text { where } x \geq-1 \\ &g(x)=\frac{1}{8}\left(x^{3 / 2}-8\right), \text { where } x \geq 0 \end{aligned} $$

Suppose that \(f\) is a one-to-one function such that \(f(2)=5\). Find \(f^{-1}(5)\).

Sketch the graph of the first function by plotting points if necessary. Then use transformation(s) to obtain the graph of the second function. \(y=x^{2}, \quad y=2 x^{2}-4 x+1\)
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