/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Sketch the graph of the function... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch the graph of the function by plotting points. $$g(x)=1+\log x$$

Short Answer

Expert verified
Sketch the graph using points \((1, 1)\), \((2, 1.301)\), and \((10, 2)\), showing a vertical shift up of the logarithmic function.

Step by step solution

01

Understand the Function

We are given the function \(g(x) = 1 + \log x\). This function is a vertical shift of the basic logarithmic function \(\log x\), shifted up by 1 unit.
02

Determine the Domain

The logarithmic function \(\log x\) is defined for \(x > 0\). Thus, the domain of \(g(x)\) is also \(x > 0\).
03

Calculate Key Points

To plot the function, calculate key points. For example: \(x = 1\), \(x = 2\), \(x = 10\). Use these values to find \(g(x)\).- For \(x = 1\):\[g(1) = 1 + \log 1 = 1 + 0 = 1\]- For \(x = 2\):\[g(2) = 1 + \log 2 \approx 1 + 0.301 = 1.301\]- For \(x = 10\):\[g(10) = 1 + \log 10 = 1 + 1 = 2\]
04

Sketch the Graph

Using the calculated points \((1, 1)\), \((2, 1.301)\), and \((10, 2)\), plot these on a graph. Draw a curve through these points smoothly. The curve should approach the y-axis asymptotically, never touching it, and should continue to rise slowly as \(x\) increases.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertical Shift
A vertical shift adjusts the entire graph of a function up or down without changing its shape. For the function \(g(x) = 1 + \log x\), we're starting with the basic logarithmic function \(\log x\). This function is shifted vertically upwards by 1 unit.

To visualize this, imagine taking the graph of \(\log x\) and lifting it so every point is one unit higher on the y-axis. Despite the shift, the overall behavior of the graph remains the same—it rises slowly as \(x\) increases and never crosses the y-axis.

The vertical shift is simply adding a constant to the function, which in our case is the number 1.
  • Original function: \(\log x\)
  • Shifted function: \(1 + \log x\)
  • Effect: Every point on the \(\log x\) graph is now one unit higher
Domain of a Function
The domain of a function includes all the input values \(x\) for which the function is defined. For logarithmic functions, it's essential to understand what values are allowed.

The function \(\log x\) is defined only for positive values of \(x\), meaning \(x > 0\). This is because logarithms of non-positive numbers are not defined in real number systems.

Similarly, for the function \(g(x) = 1 + \log x\), the domain remains \(x > 0\). The addition of 1 in our function only affects the vertical position of points, not the values \(x\) can take.
  • Logarithmic base function domain: \(x > 0\)
  • Shifted function domain: \(x > 0\) remains unchanged
Understand the domain is crucial to avoid plotting the graph in areas where it doesn't exist.
Key Points
Key points are specific values of \(x\) used to help graph a function accurately. They act like anchor points, ensuring that the curve we draw is precise.

For \(g(x) = 1 + \log x\), some useful key points include when \(x = 1\), \(x = 2\), and \(x = 10\). Here's how these are calculated:
  • \(x = 1\): \(g(1) = 1 + \log 1 = 1 + 0 = 1\)
  • \(x = 2\): \(g(2) = 1 + \log 2 \approx 1.301\)
  • \(x = 10\): \(g(10) = 1 + \log 10 = 1 + 1 = 2\)

These key points give us specific coordinates to plot on our graph, like \((1, 1)\), \((2, 1.301)\), and \((10, 2)\). By plotting these coordinates and connecting them with a smooth curve, we ensure our graph reflects the function accurately.
Asymptote
An asymptote is a line that a graph approaches but never actually reaches. In the case of logarithmic functions, these are vertical asymptotes.

For \(\log x\) and our function \(g(x) = 1 + \log x\), there is a vertical asymptote at the line \(x = 0\). The curve will get infinitely close to the y-axis as \(x\) approaches zero from the positive side but will never touch or cross the axis.

Understanding asymptotes is important for comprehending the function's behavior near certain values. Knowing that the function approaches the y-axis without crossing gives us expectations about how to sketch the curve.
  • Logarithmic function asymptote: \(x = 0\)
  • Behavior: The function nears \(x = 0\) but does not cross it

This characteristic is what helps define the shape and limits of the graph, ensuring that despite any vertical shifts, the asymptotic behavior remains the same.

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Most popular questions from this chapter

Compare log 1000 to the number of digits in 1000 . Do the same for 10,000 . How many digits does any number between 1000 and 10,000 have? Between what two values must the common logarithm of such a number lie? Use your observations to explain why the number of digits in any positive integer \(x\) is \(\|\log x\|+1 .\) (The symbol \(\|n\|\) is the greatest integer function defined in Section 2.2 . ) How many digits does the number \(2^{100}\) have?

What is wrong with the following argument? $$\begin{aligned}\log 0.1 &<2 \log 0.1 \\\&=\log (0.1)^{2} \\\&=\log 0.01 \\\\\log 0.1 &<\log 0.01 \\ 0.1 &<0.01\end{aligned}$$

Some biologists model the number of species \(S\) in a fixed area \(A\) (such as an island) by the speciesarea relationship $$\log S=\log c+k \log A$$ where \(c\) and \(k\) are positive constants that depend on the type of species and habitat. (a) Solve the equation for \(S\). (b) Use part (a) to show that if \(k=3,\) then doubling the area increases the number of species eightfold.

Draw graphs of the family of functions \(y=\log _{a} x\) for \(a=2, e\) \(5,\) and 10 on the same screen, using the viewing rectangle \([0,5]\) by \([-3,3]\). How are these graphs related?

These exercises deal with logarithmic scales. Inverse Square Law for Sound A law of physics states that the intensity of sound is inversely proportional to the square of the distance \(d\) from the source: \(I=k / d^{2}\) (a) Use this model and the equation $$B=10 \log \frac{I}{I_{0}}$$ (described in this section) to show that the decibel levels \(B_{1}\) and \(B_{2}\) at distances \(d_{1}\) and \(d_{2}\) from a sound source are related by the equation $$B_{2}=B_{1}+20 \log \frac{d_{1}}{d_{2}}$$ (b) The intensity level at a rock concert is \(120 \mathrm{dB}\) at a distance \(2 \mathrm{m}\) from the speakers. Find the intensity level at a distance of \(10 \mathrm{m}\).

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