Question

Answer the questions about each function. $$f(x)=\frac{2 x}{x-2}$$ (a) Is the point \(\left(\frac{1}{2},-\frac{2}{3}\right)\) on the graph of \(f ?\) (b) If \(x=4,\) what is \(f(x) ?\) What point is on the graph of \(f ?\) (c) If \(f(x)=1,\) what is \(x ?\) What point(s) are on the graph of \(f ?\) (d) What is the domain of \(f ?\) (e) List the \(x\) -intercepts, if any, of the graph of \(f\). (f) List the \(y\) -intercept, if there is one, of the graph of \(f\).

Short answer

(a) Yes, (b) 4; (4, 4), (c) x = -2; (-2, 1), (d) \((-\infty, 2) \cup (2, \infty)\), (e) (0, 0), (f) (0, 0).

Step-by-step solution

Determine if the point \(\frac{1}{2}, -\frac{2}{3}\) is on the graph

Substitute \(x = \frac{1}{2}\) into the function \(f(x) = \frac{2x}{x-2}\) and see if \(f(\frac{1}{2}) = -\frac{2}{3}\).\[f\left(\frac{1}{2}\right) = \frac{2 \cdot \frac{1}{2}}{\frac{1}{2} - 2} = \frac{1}{-\frac{3}{2}} = -\frac{2}{3}\]Therefore, the point \(\frac{1}{2}, -\frac{2}{3}\) is on the graph.

Calculate \(f(4)\) and determine the corresponding point

Substitute \(x = 4\) into the function \(f(x)\):\[f(4) = \frac{2\cdot 4}{4-2} = \frac{8}{2} = 4\]The point on the graph is \((4, 4)\).

Find \(x\) if \(f(x) = 1\)

Set \(f(x) = 1\) and solve for \(x\):\[1 = \frac{2x}{x-2}\]Multiply both sides by \(x-2\):\[x-2 = 2x \rightarrow x - 2x = 2 \rightarrow -x = 2 \rightarrow x = -2\]Thus, the point is \((-2, 1)\).

Determine the domain of \(f\)

The function \(f(x)\) is undefined when the denominator is zero:\[x - 2 = 0 \rightarrow x = 2\]Therefore, the domain is \( (-\infty, 2) \cup (2, \infty) \).

Find the x-intercepts

Set \(f(x) = 0\) and solve for \(x\):\[0 = \frac{2x}{x-2}\]Since the numerator must be zero, solve \(2x = 0\):\[x = 0\]So, there is an x-intercept at \((0, 0)\).

Find the y-intercept

Substitute \(x = 0\) into the function \(f(x)\):\[f(0) = \frac{2\cdot 0}{0-2} = 0\]Therefore, the y-intercept is at \((0, 0)\).

Key concepts

Function Evaluation

Function evaluation is the process of finding the value of a function for a given input. In other words, we substitute a specific value for the function's variable and perform the operations indicated to find the output. For instance, to evaluate the function \( f(x) = \frac{2x}{x-2} \) at \( x = 4 \), we substitute 4 in place of \( x \) and solve:

\[ f(4) = \frac{2\cdot 4}{4-2} = \frac{8}{2} = 4 \]
This tells us that the output is 4 when the input is 4.

Function evaluation can also verify if a point is on a function's graph. For example, to check if the point \( \left( \frac{1}{2}, -\frac{2}{3} \right) \) is on the graph of \( f(x) = \frac{2x}{x-2} \), substitute \( x = \frac{1}{2} \) into the function:

\[ f\left( \frac{1}{2} \right) = \frac{2 \cdot \frac{1}{2}}{\frac{1}{2} - 2} = \frac{1}{-\frac{3}{2}} = -\frac{2}{3} \]
Since this equals the y-value of the point, we know the point is on the graph.

Domain and Range

The domain of a function is all the input (x) values for which the function is defined, while the range is all the possible output (y) values. For rational functions like \( f(x) = \frac{2x}{x-2} \), the denominator cannot be zero, as division by zero is undefined.

To find the domain, we identify the x-values that make the denominator zero and exclude them:

\[ x - 2 = 0 \rightarrow x = 2 \]
Therefore, the domain is:

\[ (-\infty, 2) \cup (2, \infty) \]

The range of \( f(x) \) usually requires further analysis, but in rational functions, certain values might make the function undefined. Here, \( f(x) \) has no restrictions on the range based on the simplified analysis.

Intercepts

Intercepts are the points where the graph of a function crosses the axes. There are x-intercepts and y-intercepts.

X-Intercept: To find x-intercepts, set the function equal to zero and solve for x. For \( f(x) = \frac{2x}{x-2} \), set \( f(x) = 0 \):

\[ 0 = \frac{2x}{x-2} \]
This implies that the numerator must be zero:

\[ 2x = 0 \rightarrow x = 0 \]
Thus, the x-intercept is at (0, 0).

Y-Intercept: To find the y-intercept, set \( x = 0 \) and solve for \( f(0) \):

\[ f(0) = \frac{2 \cdot 0}{0-2} = 0 \]
So, the y-intercept is also at (0, 0).

Intercepts provide crucial points that help in graphing the function and understanding its behavior.