Problem 10
Question
In Exercises 9–16, sketch the graph of the function and state its domain. $$ f(x)=-2 \ln x $$
Step-by-Step Solution
Verified Answer
The domain of the function \(f(x)=-2 \ln x\) is (0, +\infty). It is a decreasing and steep graph, starting from (0,+\infty), crossing the x-axis at (1,0) and extending to (+\infty, -\infty)
1Step 1: Understand the basic logarithm function
The basic function \(\ln x\) has a graph that crosses the x-axis at \(x=1\) and reaches to positive infinity as \(x \to +\infty\) and approaches negative infinity as \(x \to 0^+\). The domain of \(\ln x\) is \((0, +\infty)\) and the range is \(-\infty, +\infty\). The function is increasing and has a vertical asymptote at \(x=0\).
2Step 2: Apply the transformation
Applying the transformation, \(f(x)=-2 \ln x\) would be a reflection of the graph of \(\ln x\) in the x-axis owing to the negative sign and a steepening of the graph due to the factor 2. Now, the graph crosses at x=1 downwards instead of upwards since the graph is reflected. As \(x \to +\infty\), the function goes to negative infinity instead of positive and as \(x \to 0^+\), the function heads to positive infinity instead of negative. The domain remains \((0, +\infty)\), and the range is also \(-\infty, +\infty\) but now the function is decreasing instead of increasing.
3Step 3: Sketch the graph
Start by drawing a vertical asymptote at \(x=0\). Then, sketch the curve so that it crosses the x-axis at \(x=1\) and goes downwards as \(x\) increases and upwards as \(x\) approaches \(0\). The final graph is a decreasing and steep graph, starting from \((0,+\infty)\), crossing the x-axis at (1,0) and extending to \((+\infty,-\infty)\).
Key Concepts
Domain of a FunctionLogarithmic TransformationsReflections in the Coordinate Plane
Domain of a Function
The concept of a function's domain is crucial for understanding where the function is defined and usable. In the case of logarithmic functions like \(\ln x\), the domain is all positive real numbers, \( (0, +\infty)\). This means that \(\ln x\) can only take positive inputs, as the logarithm of zero and negative numbers is undefined.
In our transformed function \(f(x) = -2 \ln x\), the domain remains \( (0, +\infty)\). This is because the transformation involves multiplying by constants and reflecting, neither of which alters the input restrictions. Regardless of what transformations are applied, logarithmic functions will continue to only accept values greater than zero.
This domain significantly impacts the graph because it paints a picture of where the function starts and how it behaves. You can visualize the domain as the area along the x-axis where the graph exists. So, for \(f(x) = -2 \ln x\), the graph will only live on the positive side of the x-axis.
In our transformed function \(f(x) = -2 \ln x\), the domain remains \( (0, +\infty)\). This is because the transformation involves multiplying by constants and reflecting, neither of which alters the input restrictions. Regardless of what transformations are applied, logarithmic functions will continue to only accept values greater than zero.
This domain significantly impacts the graph because it paints a picture of where the function starts and how it behaves. You can visualize the domain as the area along the x-axis where the graph exists. So, for \(f(x) = -2 \ln x\), the graph will only live on the positive side of the x-axis.
Logarithmic Transformations
Transformations modify the appearance and position of standard function graphs. For logarithmic functions like \(f(x) = -2 \ln x\), transformations can include scaling, reflecting, or shifting the graph.
The factor of \(-2\) in \(f(x)\) plays a crucial role in how the function graph is altered. The negative sign reflects the graph across the x-axis, flipping it upside down. This means where the \(\ln x\) graph was rising to positive infinity, \(f(x)\) will instead fall to negative infinity.
The factor of \(-2\) in \(f(x)\) plays a crucial role in how the function graph is altered. The negative sign reflects the graph across the x-axis, flipping it upside down. This means where the \(\ln x\) graph was rising to positive infinity, \(f(x)\) will instead fall to negative infinity.
- Multiplying by 2 steepens the curve, making changes more pronounced as x increases.
- These changes do not affect the domain of the function, but they dramatically change the visual graph.
Reflections in the Coordinate Plane
Reflections in the coordinate plane are powerful transformations that flip a graph across one of the axes. For the function \(f(x) = -2 \ln x\), a reflection in the x-axis is applied.
Reflections change the sign of the output values. A reflection in the x-axis involves turning the graph of \(\ln x\) upside down. So, wherever there used to be positive y-values for \(\ln x\), now there are negative values for \(f(x)\).
Reflections change the sign of the output values. A reflection in the x-axis involves turning the graph of \(\ln x\) upside down. So, wherever there used to be positive y-values for \(\ln x\), now there are negative values for \(f(x)\).
- This reflection transforms increasing behavior of \(\ln x\) into decreasing behavior for \(f(x)\).
- The reflection leaves the domain of the function untouched but modifies the range.
Other exercises in this chapter
Problem 10
Evaluate the expression without using a calculator. \(\operatorname{arcsec}(-\sqrt{2})\)
View solution Problem 10
Solving an Exponential or Logarithmic Equation In Exercises 1-16, solve for \(x\) accurate to three decimal places. $$ \frac{5000}{1+e^{2 x}}=2 $$
View solution Problem 10
Finding an Indefinite Integral In Exercises \(1-26,\) find the indefinite integral.. $$ \int \frac{x^{2}-2 x}{x^{3}-3 x^{2}} d x $$
View solution Problem 11
In Exercises 7–14, verify the identity. $$ \sinh 2 x=2 \sinh x \cosh x $$
View solution