Problem 53 - College Algebra Essentials

Question

Begin by graphing \(f(x)=\log _{2} x .\) Then use transformations of this graph to graph the given function. What is the vertical asymptote? Use the graphs to determine each function's domain and range. $$g(x)=\log _{2}(x+1)$$

Step-by-Step Solution

Answer

The vertical asymptote for function \(f(x) = \log_2 x\) is at \(x = 0\), and for function \(g(x) = \log_2 (x + 1)\) is at \(x = -1\). The domain for \(f(x) = \log_2 x\) is \(x > 0\), for \(g(x) = \log_2 (x + 1)\) it is \(x > -1\). For both functions, the range is all real numbers.

Step 1: Graphing \(f(x) = \log_2 x\)

Begin by graphing the function \(f(x) = \log_2 x\). The graph of \(y = \log_2 x\) is a curve that passes through the point (1, 0) since \(2^0=1\) and increases to the right. The graph has a vertical asymptote at \(x=0\).

Step 2: Transforming the graph of \(f(x)\) to \(g(x) = \log_2 (x + 1)\)

Now, use the graph of \(f(x)\) to create that of \(g(x) = \log_2 (x + 1)\). This function is same as \(f(x)\) but shifted one unit to the left. The vertical asymptote is also shifted one unit to the left, going to \(x=-1\).

Step 3: Determining the vertical asymptote

The vertical asymptote for \(f(x) = \log_2 x\) is at \(x = 0\), and for \(g(x) = \log_2 (x + 1)\), it is \(x = -1\) due to the shift in the graph.

Step 4: Finding the domain and range

The domain is the set of all allowable values for x. For \(f(x) = \log_2 x\), the domain is \(x > 0\) and the range is all real numbers. For \(g(x) = \log_2 (x + 1)\), the domain is \(x > -1\) due to the shift, and the range remains all real numbers.