Questions: Consider the function f(x)=10/x. a. Complete the table of values below. x f(x)=10/x -1 -0.1 -0.001 0.001 0.1 1 b. Based on your work in part (a), which of the following are true? Select all that apply. As x → 0+, f(x) decreases without bound. As x → 0-, f(x) decreases without bound. As x → 0-, f(x) increases without bound. As x → 0+, f(x) increases without bound.

Consider the function f(x)=10/x.
a. Complete the table of values below.
x  f(x)=10/x
-1  
-0.1  
-0.001  
0.001  
0.1  
1  
b. Based on your work in part (a), which of the following are true? Select all that apply.
As x → 0+, f(x) decreases without bound.
As x → 0-, f(x) decreases without bound.
As x → 0-, f(x) increases without bound.
As x → 0+, f(x) increases without bound.
Transcript text: imathas.rationalreasoning.net/assessment/showtest.php Unlimited attempts. Score on last attempt: 1. Score in gradebook: 1 Consider the function $f(x)=\frac{10}{x}$. a. Complete the table of values below. \begin{tabular}{|c|c|} \hline$x$ & $f(x)=\frac{10}{x}$ \\ \hline-1 & $\square$ \\ \hline-0.1 & $\square$ \\ \hline-0.001 & $\square$ \\ \hline 0.001 & $\square$ \\ \hline 0.1 & $\square$ \\ \hline 1 & $\square$ \\ \hline \end{tabular} b. Based on your work in part (a), which of the following are true? Select all that apply. As $x \rightarrow 0^{+}, f(x)$ decreases without bound. As $x \rightarrow 0^{-}, f(x)$ decreases without bound. As $x \rightarrow 0^{-}, f(x)$ increases without bound. As $x \rightarrow 0^{+}, f(x)$ increases without bound. Submit Question 3. Points possible: 2 Unlimited attempts.
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Solution

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Solution Steps

Solution Approach

a. To complete the table of values for the function \( f(x) = \frac{10}{x} \), substitute each given \( x \) value into the function to calculate \( f(x) \).

b. Analyze the behavior of \( f(x) \) as \( x \) approaches 0 from the positive side (\( x \rightarrow 0^{+} \)) and from the negative side (\( x \rightarrow 0^{-} \)). Determine whether \( f(x) \) increases or decreases without bound in each case.

Step 1: Calculate \( f(x) \) for Given Values

We evaluate the function \( f(x) = \frac{10}{x} \) for the specified \( x \) values:

  • For \( x = -1 \): \[ f(-1) = \frac{10}{-1} = -10.0 \]

  • For \( x = -0.1 \): \[ f(-0.1) = \frac{10}{-0.1} = -100.0 \]

  • For \( x = -0.001 \): \[ f(-0.001) = \frac{10}{-0.001} = -10000.0 \]

  • For \( x = 0.001 \): \[ f(0.001) = \frac{10}{0.001} = 10000.0 \]

  • For \( x = 0.1 \): \[ f(0.1) = \frac{10}{0.1} = 100.0 \]

  • For \( x = 1 \): \[ f(1) = \frac{10}{1} = 10.0 \]

The completed table of values is: \[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -1 & -10.0 \\ -0.1 & -100.0 \\ -0.001 & -10000.0 \\ 0.001 & 10000.0 \\ 0.1 & 100.0 \\ 1 & 10.0 \\ \hline \end{array} \]

Step 2: Analyze the Behavior of \( f(x) \) as \( x \) Approaches 0

We analyze the limits of \( f(x) \) as \( x \) approaches 0 from both the positive and negative sides:

  • As \( x \rightarrow 0^{+} \): \[ f(x) \rightarrow +\infty \quad \text{(increases without bound)} \]

  • As \( x \rightarrow 0^{-} \): \[ f(x) \rightarrow -\infty \quad \text{(decreases without bound)} \]

Final Answer

The completed table of values is: \[ \begin{array}{|c|c|} \hline x & f(x) \\ \hline -1 & -10.0 \\ -0.1 & -100.0 \\ -0.001 & -10000.0 \\ 0.001 & 10000.0 \\ 0.1 & 100.0 \\ 1 & 10.0 \\ \hline \end{array} \]

The behavior of \( f(x) \) is:

  • As \( x \rightarrow 0^{+} \), \( f(x) \) increases without bound.
  • As \( x \rightarrow 0^{-} \), \( f(x) \) decreases without bound.

Thus, the answers to the multiple-choice questions are:

  • As \( x \rightarrow 0^{+}, f(x) \) increases without bound: True
  • As \( x \rightarrow 0^{-}, f(x) \) decreases without bound: True
  • As \( x \rightarrow 0^{-}, f(x) \) increases without bound: False
  • As \( x \rightarrow 0^{+}, f(x) \) decreases without bound: False

The final boxed answer is: \[ \boxed{\text{True for the first two statements, False for the last two.}} \]

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