Problem 210 - Precalculus

Question

For the following exercises, state the domain, range, and \(x\) - and \(y\) -intercepts, if they do not exist, write DNE. $$f(x)=\log (5 x+10)+3$$

Step-by-Step Solution

Answer

Domain: \(x > -2\), Range: \((-\infty, \infty)\), \(x\)-intercept: solve for value, \(y\)-intercept: 4.

Step 1: Identify the Domain

The domain of the function is determined by the expression inside the logarithm. For \( \log(5x+10) \) to be defined, \( 5x+10 > 0 \). Solving this inequality, we get \( 5x > -10 \) which simplifies to \( x > -2 \). Thus, the domain is \( x > -2 \).

Step 2: Determine the Range

The logarithmic function can take any real number as its output. Since \( \log(5x+10) \) can span all real numbers, adding 3 to it shifts the range upwards by 3 units. Therefore, the range is all real numbers \( (-\infty, \infty) \).

Step 3: Calculate the x-intercept

To find the \(x\)-intercept, set \(f(x) = 0\). Thus, \( \log(5x+10) + 3 = 0 \). Solving for \(x\), we have \( \log(5x+10) = -3 \). Converting to exponential form, \(5x+10 = 10^{-3}\), which simplifies to \(5x = 10^{-3} - 10\). Solving for \(x\), \( x = \frac{10^{-3} - 10}{5} \). Calculate this to get a number approximate. The exact value is the \(x\)-intercept.

Step 4: Determine the y-intercept

The \(y\)-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the function: \(f(0) = \log(5(0)+10)+3 = \log(10) + 3 = 1 + 3 = 4\). Therefore, the \(y\)-intercept is 4.

Key Concepts

Domain

For a logarithmic function like \(f(x) = \log(5x+10) + 3\), the domain is the set of possible \(x\)-values that we can input into the function and still get a real number as an output. A logarithm, \(\log(a)\), is defined only when \(a > 0\). Therefore, inside the parentheses, \(5x + 10\) must be positive.
To find when \(5x + 10 > 0\), we solve the inequality:

• Solve \(5x + 10 > 0\).
• Subtract 10 from both sides to get \(5x > -10\).
• Divide by 5 to find \(x > -2\).

This tells us that any \(x\) greater than \(-2\) will work in this function, meaning the domain of \(f(x)\) is \(x > -2\).
This is an important step, as inputting values outside of this domain would give us an "undefined" or a nonexistent output from the logarithmic function.

Range

The range of a function refers to all the possible output values (\(y\)-values) it can produce. For the function \(f(x) = \log(5x+10) + 3\), understanding the range involves recognizing that logarithmic functions can take any real value as output.
The basic logarithmic function, \(\log(x)\), spans any real number from negative infinity to positive infinity as its output when the domain is satisfied. Our function involves adding 3, which shifts the graph vertically upward by 3 units.

• Start with outputs \((-\infty, \infty)\) for \(\log\).
• Add 3 to all outputs to get \((-\infty, \infty)\) shifted 3 units up.

Therefore, the range of \(f(x)\) is \((-\infty, \infty)\) as well. This upward shift doesn't restrict the range since all real outputs are still possible.

x-intercepts

To find \(x\)-intercepts, which are points where the graph crosses the \(x\)-axis, we need to set the function \(f(x)\) to zero and solve for \(x\). For the function \(f(x) = \log(5x+10) + 3\), this means:

• Set \(\log(5x+10) + 3 = 0\).
• Solve for \(\log(5x+10) = -3\).

By converting the logarithmic equation to its exponential form, \(5x+10 = 10^{-3}\), we get a solvable equation:

• Subtract 10 from both sides to get \(5x = 10^{-3} - 10\).
• Divide by 5 to isolate \(x\).
• Now, calculate the exact \(x\) value, this is your \(x\)-intercept.

This spot is the particular \(x\) value where our function touches the \(x\)-axis, indicating the function's zero point.

y-intercepts

The \(y\)-intercept is where the graph of a function crosses the \(y\)-axis, which occurs when \(x = 0\). For our function \(f(x) = \log(5x + 10) + 3\), to find this point:

• Substitute \(x = 0\) into the function: \(f(0) = \log(5(0) + 10) + 3\).
• This simplifies to \(f(0) = \log(10) + 3\).
• Since \(\log(10) = 1\), this becomes \(f(0) = 1 + 3 = 4\).

The \(y\)-intercept is therefore 4. This simple substitution gives us the exact point the function crosses the \(y\)-axis, which can be very helpful for sketching graphs or understanding how the function behaves at the starting point.