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Chapter 4: Problem 55

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. $$h(x)=1+\log _{2} x$$

Short Answer

Expert verified
The vertical asymptote for both functions is at \(x=0\). The domain of both functions is \(x > 0\) and the range of \(f(x) = \log_2x\) is all real numbers. However, the range of \(h(x) = 1 + \log_2x\) is \(y > 1\).

Step by step solution

01

- Graphing the Function \(f(x) = \log_2x\)

Start with the basic function \(f(x) = \log_2x\). It is important to recall that the basic graph of a logarithmic function \(f(x) = \log_bx\) has a vertical asymptote at \(x=0\), passes through the point \((1, 0)\), and increases without bound as \(x\) increases. Draw the graph appropriately on the 2-dimensional plane according to these characteristics.
02

- Graphing \(h(x) = 1 + \log_2x\)

The function \(h(x) = 1 + \log_2x\) can be obtained from \(f(x) = \log_2x\) by shifting the graph of \(f(x)\) upward by one unit. This is because adding a constant to a function causes a vertical shift in the graph of the function. Thus, the graph of \(h(x)\) will be the same as the graph of \(f(x)\), but moved upward by one unit.
03

- Determining the Vertical Asymptote

The vertical asymptote for both functions \(f(x) = \log_2x\) and \(h(x) = 1 + \log_2x\) is at \(x=0\). This is true even after the transformation because vertical transformations do not affect the positioning of the vertical asymptote, they only shift the graph up or down.
04

- Determining the Domain and Range

The domain of both functions is \(x > 0\) since the logarithmic functions are only defined for positive values of \(x\). The range of \(f(x) = \log_2x\) is all real numbers because the logarithm of any positive number can take any real value. However, the range of \(h(x) = 1 + \log_2x\) is translated to \(y > 1\), because of the transformation. The graph of \(h(x)\) is shifted up so all the \(y\) values are greater than 1.

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Most popular questions from this chapter

One problem with all exponential growth models is that nothing can grow exponentially forever. Describe factors that might limit the size of a population.

Check each proposed solution by direct substitution or with a graphing utility. $$(\log x)(2 \log x+1)=6$$

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$\text { The domain of } f(x)=\log _{2} x \text { is }(-\infty, \infty)$$.

Evaluate the indicated logarithmic expressions without using a calculator. a. Evaluate: \(\log _{3} 81\) b. Evaluate: \(2 \log _{3} 9\) c. What can you conclude about $$\log _{3} 81, \text { or } \log _{3} 9^{2} ?$$

Explain how to use the graph of \(f(x)=2^{x}\) to obtain the graph of \(g(x)=\log _{2} x\).
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