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Chapter 3: Problem 84

Use a graphing utility to graph the function. Be sure to use an appropriate viewing window. $$f(x)=3 \ln x-1$$

Short Answer

Expert verified
The graph of \(f(x) = 3 \ln x - 1\) is increasing, with a vertical asymptote at \(x=0\). It passes through the point (1,2) and approaches negative infinity as \(x\) approaches zero from the right while it increases without bound as \(x\) approaches infinity.

Step by step solution

01

Understand the Function and the Viewing Window

The given function is \(f(x) = 3 \ln x - 1\), a logarithmic function. The function is defined for all \(x > 0\). Therefore, the appropriate viewing window for this function on the x-axis is any value greater than zero. On the y-axis, the viewing window will depend on the range of the function, which in this case is all real numbers. To begin with, a reasonable range may be from -10 to 10.
02

Use the Graphing Utility

Enter the function \(f(x) = 3 \ln x - 1\) into the graphing utility. Make sure to set the viewing window suitably. An appropriate initial window might be \(x\) in (0, 10) and \(y\) in (-10, 10).
03

Analyze the Graph

Observe the resulting graph. The graph of the function should pass through the point (1,2) because \(f(1) = 3 \ln 1 - 1 = 2\). Since the coefficient of \(\ln x\) is positive, the function is increasing. It will approach negative infinity as x approaches 0 from the right, and it will increase without bound as \(x\) approaches infinity. The function does not have a maximum or minimum value, but there is a vertical asymptote at \(x=0\).

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Most popular questions from this chapter

Write two or three sentences stating the general guidelines that you follow when solving (a) exponential equations and (b) logarithmic equations.

Find the exponential model \(y=a e^{b x}\) that fits the points shown in the graph or table. $$ \begin{array}{|l|l|l|} \hline x & 0 & 4 \\ \hline y & 5 & 1 \\ \hline \end{array} $$

Solve the equation algebraically. Round the result to three decimal places. Verify your answer using a graphing utility. $$\frac{1-\ln x}{x^{2}}=0$$

An exponential growth model has the form ________ and an exponential decay model has the form ________.

At 8: 30 A.M., a coroner was called to the home of a person who had died during the night. In order to estimate the time of death, the coroner took the person's temperature twice. At 9: 00 A.M. the temperature was \(85.7^{\circ} \mathrm{F}\), and at 11: 00 A.M. the temperature was \(82.8^{\circ} \mathrm{F}\). From these two temperatures, the coroner was able to determine that the time elapsed since death and the body temperature were related by the formula $$t=-10 \ln \frac{T-70}{98.6-70}$$ where \(t\) is the time in hours elapsed since the person died and \(T\) is the temperature (in degrees Fahrenheit) of the person's body. (This formula is derived from a general cooling principle called Newton's Law of Cooling. It uses the assumptions that the person had a normal body temperature of \(98.6^{\circ} \mathrm{F}\) at death, and that the room temperature was a constant \(70^{\circ} \mathrm{F}\).) Use the formula to estimate the time of death of the person.
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