Problem 212 - Precalculus

Question

For the following exercises, state the domain, range, and \(x\) - and \(y\) -intercepts, if they do not exist, write DNE. $$f(x)=\log _{2}(x+2)-5$$

Step-by-Step Solution

Answer

Domain: \((-2, \infty)\); Range: \((-\infty, \infty)\); \(x\)-intercept: \((30, 0)\); \(y\)-intercept: \((0, -4)\).

Step 1: Determine the Domain

The domain of the function is the set of all values of \(x\) that make the function defined. For \(f(x) = \log_2(x+2) - 5\), the argument of the logarithm \((x + 2)\) must be greater than zero. So, set \(x + 2 > 0\). Solving for \(x\), we get \(x > -2\). Thus, the domain of \(f(x)\) is \((-2, \infty)\).

Step 2: Determine the Range

The range of a logarithmic function transformed by a vertical shift involves all real numbers because the logarithmstretches infinitely in both directions as its input approaches its domain boundaries. Here \(f(x)\) is translated downwards by 5 units. Therefore, the range of \(f(x)\) is still \((- ext{infinity}, ext{infinity})\).

Step 3: Find the x-intercept

The \(x\)-intercept occurs where \(f(x) = 0\). Set \(\log_2(x+2) - 5 = 0\) and solve for \(x\). First, add 5 to both sides to get \(\log_2(x+2) = 5\). Converting from logarithmic to exponential form: \(x + 2 = 2^5\). Simplifying gives \(x + 2 = 32\), so \(x = 30\). Thus, the \(x\)-intercept is at \((30, 0)\).

Step 4: Find the y-intercept

The \(y\)-intercept occurs where \(x=0\). Substitute \(x = 0\) into the function: \(f(0) = \log_2(0+2) - 5\). Simplifying gives \(f(0) = \log_2(2) - 5\). Since \(\log_2(2) = 1\), we have \(f(0) = 1 - 5 = -4\). Thus, the \(y\)-intercept is \((0, -4)\).

Key Concepts

Domain and Range of Logarithmic Functions

In mathematics, understanding the domain and range of logarithmic functions is fundamental. For the function given, \(f(x) = \log_2(x+2) - 5\), let's break down these components:
The **domain** of a logarithmic function consists of all real numbers for which the function is defined. Remember, the argument of the logarithm must be greater than zero. Therefore, for \(\log_2(x+2)\):\[x + 2 > 0\]
When you solve for \(x\), you find \(x > -2\). This gives us the domain \((-2, \infty)\), meaning any number greater than \(-2\) can be plugged into the function.
The **range**, however, is quite different. Logarithmic functions can translate infinitely in the vertical direction because their y-values can cover any real number, especially since we've adjusted our function by subtracting 5. This vertical translation ensures that the range is all real numbers \, \((-\infty, \infty)\). This means for any real number output, there is some input that will produce it.

x-intercepts and y-intercepts in Logarithmic Functions

Interpreting intercepts, both x and y, on the graph of a logarithmic function is crucial for understanding how these functions behave.
**x-intercept**: This is where the function crosses the x-axis. For our function \(f(x) = \log_2(x+2) - 5\), set \(f(x) = 0\) and solve. So, our equation becomes \(\log_2(x+2) = 5\). By changing this to exponential form, we get \(x + 2 = 2^5\). Solving gives \(x + 2 = 32\), hence \(x = 30\). This gives us an x-intercept at the point \((30, 0)\).
**y-intercept**: This point occurs where the function crosses the y-axis, at \(x=0\). By substituting \(x = 0\) into the function, we calculate \(f(0) = \log_2(2) - 5\). Since \(\log_2(2) = 1\), we have \(f(0) = 1 - 5 = -4\), resulting in a y-intercept at \((0, -4)\).
Knowing these intercepts provides a sketch of where the graph crosses the x and y axes, aiding in better visualization.

Converting Logarithmic to Exponential Form

To deeply grasp logarithmic functions, you need to understand the conversion between logarithmic and exponential forms.
**Logarithmic Form**: Let's revisit the condition \(\log_2(x+2) = 5\), representing the x-intercept scenario. In logarithmic notation, \(\log_b(a) = c\) implies that base \(b\) raised to power \(c\) equals \(a\).
**Exponential Form**: Converting to exponential form involves the same idea. From \(\log_2(x+2) = 5\), we deduce that \(x+2 = 2^5\).
This transformation allows us to solve for \(x\) more easily:

• Calculate \(2^5\) to obtain 32.
• Solve for \(x\) by isolating it, thus \(x = 32 - 2\).
• This yields \(x = 30\).

Understanding this conversion is vital not just for solving, but for comprehending the inherent relationship between exponential growth and logarithms.