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Chapter 0: Problem 23

Find the domain of the function. $$ f(x)=\frac{\sqrt{x+2}+\sqrt{2-x}}{x^{3}-x} $$

Short Answer

Expert verified
The domain of the function \(f(x)=\frac{\sqrt{x+2}+\sqrt{2-x}}{x^{3}-x}\) is \(Domain(f(x)) = \left[-2, -1\right) \cup \left(-1, 0\right) \cup \left(0, 1\right) \cup \left(1, 2\right]\).

Step by step solution

01

Identifying problematic square roots

First, let's identify the square roots that may cause issues in the function. The square root of a negative number is not defined for real numbers. Thus, we need to consider when the expressions under the square root symbols are greater than or equal to zero: \(x+2 \geq 0\) and \(2-x \geq 0\)
02

Solve the inequalities for x

Solving the inequalities from step 1, to determine the x values which make both square roots valid: For \(x + 2 \geq 0\): Subtract 2 from both sides: \(x \geq -2\) For \(2 - x \geq 0\): Add x to both sides and subtract 2: \(x \leq 2\) By combining these inequalities, we get: \(-2 \leq x \leq 2\)
03

Identifying potential division by zero

Next, we need to consider when the denominator is equal to zero, as division by zero is undefined. To do this, set the denominator equal to zero and solve for x: \(x^3 - x = 0\)
04

Factor and solve for x

In order to find the x values that result in division by zero in the denominator, we can factor the equation from step 3. Factoring, we get: \(x(x^2 - 1) = 0\) Now we have a product equal to zero, so one of the factors must be zero: \(x = 0\) or \(x^2 - 1 = 0\) For \(x^2 - 1 = 0\), we can add 1 to both sides and take the square root: \(x = \pm 1\) So, the denominator is zero when \(x = -1, 0,\) or \(1\).
05

Removing undefined x values from the domain

Since the denominator is zero when \(x = -1, 0,\) or \(1\), these x values are not in the domain of the function. So, we must remove them from the range \(-2 \leq x \leq 2\) that we found earlier: \(-2 \leq x < -1\) or \(-1 < x < 0\) or \(0 < x < 1\) or \(1 < x \leq 2\)
06

Write the domain of the function

Finally, we can present the domain of the function as a union of the intervals from Step 5: \( Domain(f(x)) = \left[-2, -1\right) \cup \left(-1, 0\right) \cup \left(0, 1\right) \cup \left(1, 2\right] \)

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Most popular questions from this chapter

Plot the graph of the function \(f\) in an appropriate viewing window. (Note: The answer is not unique.) $$ f(x)=x^{2} \sin \frac{1}{x} $$

Let \(f(x)=x+\frac{1}{100} \sin 100 x\) a. Plot the graph of \(f\) using the viewing window \([-10,10] \times[-10,10]\) b. Plot the graph of \(f\) using the viewing window \([-0.1,0.1] \times[-0.1,0.1]\) c. Explain why the two displays obtained in parts (a) and (b) taken together give a complete description of the graph of \(f\).

a. Show that if a function \(f\) is defined at \(-x\) whenever it is defined at \(x\), then the function \(g\) defined by \(g(x)=f(x)+f(-x)\) is an even function and the function \(h\) defined by \(h(x)=f(x)-f(-x)\) is an odd function. b. Use the result of part (a) to show that any function \(f\) defined on an interval \((-a, a)\) can be written as a sum of an even function and an odd function. c. Rewrite the function $$ f(x)=\frac{x+1}{x-1} \quad-1

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

Plot the graph of the function \(f\) in an appropriate viewing window. (Note: The answer is not unique.) $$ f(x)=\frac{1}{2} \sin 2 x+\cos x $$
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