Problem 30 - Calculus for Biology and Medicine

Question

Explain how the following functions can be obtained fron \(y=\ln x\) by basic transformations: (a) \(y=\ln (1-x)\) (b) \(y=\ln (2+x)-1\) (c) \(y=-\ln (2-x)+1\)

Step-by-Step Solution

Answer

(a) Reflect across y-axis. (b) Shift left 2 and down 1. (c) Reflect across x-axis, shift right 2 and up 1.

Step 1: Original Function

The original function we are starting with is the natural logarithm defined as \( y = \ln x \). This is our base function.

Step 2: Transformation 1: Reflection Across the Y-axis

To obtain the function \( y = \ln(1-x) \), we replace \( x \) with \( 1-x \) in the original function. This is equivalent to reflecting \( y = \ln x \) across the line \( x = 0.5 \), resulting in a horizontal reflection that moves the graph to the left.

Step 3: Transformation 2: Horizontal Shift

For the function \( y = \ln (2+x) - 1 \), first replace \( x \) with \( x + 2 \) in \( y = \ln x \) to shift the function left by 2 units, resulting in \( y = \ln(x+2) \). Then, subtract 1 to shift the entire graph down by 1 unit.

Step 4: Transformation 3: Reflection and Vertical Shift

In the function \( y = -\ln (2-x) + 1 \), first replace \( x \) with \( 2-x \) to achieve a horizontal reflection and shift the graph right by 2 units. This gives you \( y = \ln(2-x) \). Multiply by \(-1\) to reflect across the x-axis, resulting in \( y = -\ln(2-x) \). Finally, add 1 to shift the graph up by 1 unit.

Key Concepts

Natural Logarithm

The natural logarithm, represented as \( \ln x \), is a fundamental mathematical function. It's used to answer the question: "To what power do we need to raise \( e \), the base of natural logarithms, to get x?" Here, \( e \) is approximately 2.71828 and is an irrational number. The domain of \( \ln x \) is \( x > 0 \), meaning it only accepts positive real numbers. This makes sense because you can't raise \( e \) to any power to get 0 or a negative number.
The graph of \( \ln x \) starts from negative infinity when \( x \) approaches zero and smoothly rises, crossing the y-axis at \( x = 1 \) where \( \ln 1 = 0 \). Understanding the natural logarithm is essential when you apply transformations like reflections and shifts, as these operations modify how the graph of the function appears.

Reflection

Reflections in mathematics alter the position of a graph without changing its size or shape. This is like flipping a page in a book over or a mirror image. When reflecting a function, it is usually done over either an axis or some specific line. Here are two primary types applicable to functions derived from \( y = \ln x \):

• Reflection Over the Y-Axis: To reflect the graph over the y-axis, every x-coordinate in the function is replaced by its negative, resulting in \( y = \ln(-x) \). However, in the problem example \( y = \ln(1-x) \), it shifts horizontally first before reflection.
• Reflection Over the X-Axis: To reflect a graph over the x-axis, you multiply the function by \(-1\). For instance, \( y = -\ln x \) is the reflection of \( \ln x \) across the x-axis, flipping it upside down.

The exercise versions involve a combination of these, achieved by replacing the variable in a specific manner to execute a horizontal transformation, and optionally applying scalars.

Horizontal Shift

A horizontal shift moves every point of a graph left or right through translation. It's performed by adding or subtracting a number directly to the x-variable. For instance, in \( y= \ln(x+2) \), the function \( \ln x \) moves to the left by 2 units since each x-value is effectively offset by 2 units to achieve the same output as before.
This means that if initially, \( x=3 \) was the input, after the transformation, \( x=1 \) provides the same value in \( \ln(x+2) \). A key point is to consider whether to shift leftwards (adding to x) or rightwards (subtracting from x). Recognize that transforming functions like this allows us to control where features such as asymptotes or maximum turn toward different x-values.

Vertical Shift

Vertical shifts adjust the position of a graph up or down and are accomplished by directly adding or subtracting a constant to the function's output. Using the function \( y = \ln x - 1 \) as an example, each output is adjusted downward by one unit.
This simple operation affects the entire graph uniformly. When you apply a vertical shift, the graph shape remains the same. There are no x-value changes, only y-values are offset. For instance, peaks, troughs, or midpoints maintain their horizontal position while sliding up or down to new vertical levels.

• Moving Down: Subtract from the function, as seen in \( y= \ln x - C\).
• Moving Up: Add to the function, exemplifying \( y= \ln x + C\).

These adjustments allow graph positioning to match specific real-world scenarios or interface more usefully with other algebraic processes.