Let's go through the problem step by step.

The function given is:

$f(x) = \frac{|x+1|}{x}$

Part (a): Fill in the table of values.

To compute $f(x)$ for different values of $x$, we calculate $f(x) = \frac{|x+1|}{x}$ for each $x$ in the table. The student has already attempted filling out the table. Let's verify:

For each $x$:

1. $x = -5$: $f(-5) = \frac{|(-5)+1|}{-5} = \frac{|-4|}{-5} = \frac{4}{-5} = -0.8$ So, $f(-5) = -0.8$.

2. $x = -4$: $f(-4) = \frac{|(-4)+1|}{-4} = \frac{|-3|}{-4} = \frac{3}{-4} = -0.75$ So, $f(-4) = -0.75$.

3. $x = -3$: $f(-3) = \frac{|(-3)+1|}{-3} = \frac{|-2|}{-3} = \frac{2}{-3} \approx -0.67$ The student wrote $-0.66$, which is an approximation. So, $f(-3) \approx -0.67$.

4. $x = -2$: $f(-2) = \frac{|(-2)+1|}{-2} = \frac{|-1|}{-2} = \frac{1}{-2} = -0.5$ So, $f(-2) = -0.5$.

5. $x = -1$: $f(-1) = \frac{|(-1)+1|}{-1} = \frac{|0|}{-1} = 0$ So, $f(-1) = 0$.

6. $x = 1$: $f(1) = \frac{|1+1|}{1} = \frac{2}{1} = 2$ The student wrote $2$, so that's correct.

7. $x = 2$: $f(2) = \frac{|2+1|}{2} = \frac{3}{2} = 1.5$ So, $f(2) = 1.5$.

8. $x = 3$: $f(3) = \frac{|3+1|}{3} = \frac{4}{3} \approx 1.33$ So, $f(3) \approx 1.33$.

9. $x = 4$: $f(4) = \frac{|4+1|}{4} = \frac{5}{4} = 1.25$ The student wrote $1.25$, which is correct.

10. $x = 5$: $f(5) = \frac{|5+1|}{5} = \frac{6}{5} = 1.2$ So, $f(5) = 1.2$.

Part (b): Are there any values of $x$ that cannot be inputted into $f(x)$?

Yes, $x = 0$ cannot be inputted into the function because it would result in division by zero. The expression $\frac{|x+1|}{x}$ is undefined when $x = 0$.

Part (c): Sketch a graph of $f(x)$.

The graph of $f(x) = \frac{|x+1|}{x}$ can be sketched based on the following observations:

• For $x > 0$, the function behaves like $f(x) = \frac{x+1}{x} = 1 + \frac{1}{x}$.
• For $x < 0$, the absolute value affects the numerator, giving $f(x) = \frac{-x-1}{x} = -1 - \frac{1}{x}$.
• There is a vertical asymptote at $x = 0$ since the function is undefined there.

To sketch the graph, plot the points from the table and note the behavior of the function approaching $x = 0$. The function increases toward $+\infty$ as $x \to 0^+$, and decreases toward $-\infty$ as $x \to 0^-$.

Would you like me to draw the graph for you or provide further details?

Relative Questions:

1. How does the absolute value in the function affect the graph?
2. What is the behavior of $f(x)$ as $x \to \infty$ and $x \to -\infty$?
3. Why does the function have a vertical asymptote at $x = 0$?
4. How would the graph change if there was no absolute value?
5. Can you determine any horizontal asymptotes for this function?

Tip:

When graphing rational functions, always check for vertical asymptotes (where the denominator is zero) and horizontal asymptotes to guide the shape of the curve.