Problem 90 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the domain, vertical asymptote, and \(x\) -intercept of the logarithmic function, and sketch its graph by hand. Verify using a graphing utility. $$h(x)=\ln (x+1)$$

Step-by-Step Solution

Answer

The domain of \(h(x)=\ln (x+1)\) is \((-1, \infty)\). The vertical asymptote is \(x=-1\). The \(x\)-intercept is \(x=0\). The function increases slowly for \(x > -1\), and approaches negative infinity as \(x\) approaches -1 from the right. The graph passes through the point \((0,0)\).

Step 1: Find the Domain

The function \(h(x) = \ln(x+1)\) is a logarithmic function. A logarithmic function is undefined when its argument \(x+1\) is less than or equal to 0. So, the domain of the function is the set of \(x\) for which \(x+1 > 0\). This solves to \( x > -1 \). So, the domain of the function is the interval \((-1, \infty)\).

Step 2: Find the Vertical Asymptote

For logarithmic functions, the vertical asymptote is at the boundary of the domain, which is \(x=-1\) in this case. This is because as \(x\) approaches -1 from the right, the function approaches negative infinity.

Step 3: Find the \(x\)-intercept

The \(x\)-intercept of a function is the value of \(x\) at which the function equals 0. This happens when the argument of the logarithm equals 1, because \(\ln 1 = 0\). Therefore, for \(h(x) = \ln(x+1)\), this occurs when \(x+1 = 1\) or \(x = 0\). Therefore, the \(x\)-intercept of the function is \(x=0\).

Step 4: Sketch the Graph

The graph of \(h(x) = \ln(x+1)\) is the graph of \(y=\ln x\) shifted one unit to the left. It increases slowly for \(x > -1\), and approaches negative infinity as \(x\) approaches -1 from the right. The graph passes through the point \((0,0)\)

Key Concepts

Domain of a Function

When working with functions, understanding their domain is crucial. The domain refers to all possible input values (often represented by the variable \(x\)) that make the function valid. For the logarithmic function \(h(x) = \ln(x+1)\), the argument inside the logarithm must be greater than zero. This requirement arises from the fact that logarithms of non-positive numbers are undefined in the real number system.

• To find the domain of \(h(x)\), set the argument greater than zero: \(x + 1 > 0\).
• This inequality simplifies to \(x > -1\).

So, the domain of \(h(x)\) is \((-1, \infty)\). This means the function takes all real numbers greater than \(-1\) as valid inputs. Knowing the domain helps us understand the range of \(x\)-values for which we can analyze the behavior of the function.

Vertical Asymptote

Vertical asymptotes are critical elements in the graph of a logarithmic function, indicating where the function increases or decreases without bound. For a function like \(h(x) = \ln(x+1)\), a vertical asymptote appears at the edge of its domain, where the function approaches undefined values.

• In \(h(x)\), this vertical boundary occurs at \(x = -1\), derived from the domain boundary \(x > -1\).
• As \(x\) gets closer to \(-1\) from the right, \(h(x)\) trends towards negative infinity.

This behavior is typical of logarithmic functions near their asymptotic barriers. Understanding vertical asymptotes helps predict how the function behaves as it approaches critical values, preventing errors in graph interpretation.

X-intercepts

An \(x\)-intercept is a point where a function's graph crosses the \(x\)-axis, meaning the function value or \(y\)-coordinate is zero at that point. For logarithmic functions, finding the \(x\)-intercept involves setting the function's expression equal to zero and solving for \(x\).

• For \(h(x) = \ln(x+1)\), set \(\ln(x+1) = 0\).
• Since \(\ln 1 = 0\), solve for \(x + 1 = 1\).
• Thus, \(x = 0\) becomes the \(x\)-intercept.

The point \((0, 0)\) marks where the function intersects the \(x\)-axis. Identifying \(x\)-intercepts helps for graph sketching and understanding where functions change from negative to positive.

Graphing Utilities

Utilizing graphing utilities can greatly aid in verifying the behavior of complex functions and visualizing their graphs. These tools provide accurate graphical representations based on the function's algebraic expression.

• For \(h(x) = \ln(x+1)\), graphing utilities can confirm its asymptotic behavior as \(x\) approaches \(-1\) and the slow linear increase post the \(x\)-intercept at \(x = 0\).
• They allow you to see visually how the curve shifts one unit left compared to \(y = \ln x\).

This validation step with a graphing utility ensures your manual graph sketch aligns with the function's exact behavior. A graphing tool serves as a reliable means to check your calculations and analysis.