Problem 60 - Precalculus with Calculus Previews

Question

Find the zeros of the given function. $$ f(x)=-1+e^{\left|3-\frac{1}{2} x\right|-2} $$

Step-by-Step Solution

Answer

The zeros of the function are \( x = 2 \) and \( x = 10 \).

Step 1: Understand the Function

The function given is \( f(x) = -1 + e^{|3 - \frac{1}{2}x| - 2} \). We need to find the values of \( x \) for which \( f(x) = 0 \). This means solving the equation \( -1 + e^{|3 - \frac{1}{2}x| - 2} = 0 \).

Step 2: Isolate the Exponential Term

Begin by setting the equation equal to zero: \(-1 + e^{|3 - \frac{1}{2}x| - 2} = 0\). Add 1 to both sides to isolate the exponential term: \(e^{|3 - \frac{1}{2}x| - 2} = 1\).

Step 3: Apply Logarithm Properties

Since the base of the natural exponential function is \( e \), we can equate the exponent directly to zero: \(|3 - \frac{1}{2}x| - 2 = 0\). By solving this equation, we will find the values of \( x \) for which the expression is zero.

Step 4: Solve the Absolute Value Equation

Rewrite \(|3 - \frac{1}{2}x| - 2 = 0\) as \(|3 - \frac{1}{2}x| = 2\). This splits into two separate equations: 1. \(3 - \frac{1}{2}x = 2\) 2. \(3 - \frac{1}{2}x = -2\).

Step 5: Solve the First Linear Equation

For the equation \(3 - \frac{1}{2}x = 2\), subtract 3 from both sides to get \(-\frac{1}{2}x = -1\). Multiply both sides by \(-2\) to solve for \( x \): \[ x = 2. \]

Step 6: Solve the Second Linear Equation

For the equation \(3 - \frac{1}{2}x = -2\), subtract 3 from both sides to get \(-\frac{1}{2}x = -5\). Multiply both sides by \(-2\) to solve for \(x\): \[ x = 10. \]

Step 7: Combine the Results

The solutions to the equation are \( x = 2 \) and \( x = 10 \). These are the zeros of the function \( f(x) \).

Key Concepts

Absolute Value

An absolute value refers to the distance of a number from zero on the number line, regardless of direction. For instance, both 5 and -5 have an absolute value of 5. It's often denoted by vertical bars, like this: \(|x|\). Absolute values play a crucial role when solving equations that involve distances and certain transformations.
In our function, \(|3 - \frac{1}{2}x|\), the absolute value ensures the expression inside the bars remains non-negative. This leads us to consider two separate cases:

• \(3 - \frac{1}{2}x = 2\)
• \(3 - \frac{1}{2}x = -2\)

This division into two scenarios is fundamental because it covers all possibilities provided by the absolute distillation of positive and negative deviations.

Exponential Function

The exponential function is characterized by the formula \(e^x\), where \(e\) is the base of the natural logarithm. It's approximately equal to 2.71828 and has unique properties that make it essential in growth and decay processes.
In the context of our exercise, the function \(f(x) = -1 + e^{|3 - \frac{1}{2}x| - 2}\) involves an exponential term. Typically, exponential functions rapidly increase or decrease, depending on whether the power is positive or negative.
To simplify the function, isolating the exponential term is crucial. We do this by setting the entire equation to zero, which translates into manipulating the exponential expression towards an equation with manageable arithmetic, leading to simpler linear forms.

Solving Equations

Solving equations involves finding the value(s) of the variable(s) that make the equation true. In mathematical problems, like finding the zeros of a function, solving the equation correctly is essential.
For the given function, this means solving the equation \(-1 + e^{|3 - \frac{1}{2}x| - 2} = 0\). By isolating the exponential part and using properties of logarithms, we turn this problem into simpler, more straightforward cases:

• First, we solve \(3 - \frac{1}{2}x = 2\) resulting in \(x = 2\).
• Secondly, \(3 - \frac{1}{2}x = -2\) which gives us \(x = 10\).

Combining these solutions helps us pinpoint the exact zeros of the function, which are the real solutions where the function intersects the x-axis.

Natural Logarithm

The natural logarithm is the inverse function of the exponential function and is denoted as \ln(x)\. It is powerful in solving equations where the variable is in an exponent.
While explicitly not needed to fully solve this particular exercise, understanding the natural logarithm helps in similar situations. When solving equations that have forms like \(e^{expression} = constant\), we often take the natural logarithm of both sides to bring down the exponent:
Generally, for \(e^x = a\), we use \ln(e^x) = x\. Consequently, \ln(a)\ becomes the exponent's value, simplifying directly to additional layers of arithmetic or algebraic manipulation.
Leveraging these properties of natural logarithms simplifies complex exponential expressions and aids in revealing solutions more simply and elegantly.