Problem 75 - Precalculus with Calculus Previews

Question

Find the points on the graph of the given function that have the indicated \(y\) -coordinate. $$ f(x)=1-e^{-x^{2}} ; \quad \frac{1}{2} $$

Step-by-Step Solution

Answer

The points are \((\sqrt{\ln(2)}, \frac{1}{2})\) and \((-\sqrt{\ln(2)}, \frac{1}{2})\)."

Step 1: Set the given function equal to the y-coordinate

Start by setting the given function equal to the specified y-coordinate. The function is \( f(x) = 1 - e^{-x^2} \) and the y-coordinate provided is \(\frac{1}{2}\). This gives us the equation: \[ 1 - e^{-x^2} = \frac{1}{2} \]

Step 2: Solve for the exponential term

Subtract 1 from both sides to begin isolating the exponential term:\[-e^{-x^2} = \frac{1}{2} - 1\]\[-e^{-x^2} = -\frac{1}{2}\]Now multiply both sides by \(-1\) to obtain:\[e^{-x^2} = \frac{1}{2}\]

Step 3: Take the natural logarithm of both sides

Apply the natural logarithm function to both sides to solve for \(x^2\):\[-x^2 = \ln\left(\frac{1}{2}\right)\]

Step 4: Solve for x^2

Multiply both sides by \(-1\) to get:\[x^2 = -\ln\left(\frac{1}{2}\right)\]

Step 5: Simplify the expression by using logarithmic identities

Simplify the argument using the property \(\ln\left(\frac{1}{2}\right) = -\ln(2)\), changing our equation to:\[x^2 = -(-\ln(2)) = \ln(2)\]

Step 6: Solve for x

Take the square root of both sides to find \(x\):\[x = \pm \sqrt{\ln(2)}\]This indicates there are two points on the graph, one for each of \(x = \sqrt{\ln(2)}\) and \(x = -\sqrt{\ln(2)}\).

Key Concepts

Natural Logarithm

In mathematics, the natural logarithm is a special logarithm with base \(e\). The constant \(e\) approximately equal to 2.71828, is an irrational number that arises naturally in many areas of mathematics. It represents the constant rate of growth shared by all continually growing processes.

Natural logarithms, denoted as \(\ln(x)\), have fascinating properties:

• \(\ln(1) = 0\), since \(e^0 = 1\).
• \(\ln(e) = 1\), as \(e^1 = e\).
• Inverse function: \(e^x\) and \(\ln(x)\) cancel each other out.

In our context, logarithms help solve equations involving exponential terms. By taking the logarithm on both sides, you can strip away the exponential to solve for the variable embedded within. Understanding these properties is crucial when you're manipulating equations like the one in the exercise: \(e^{-x^2} = \frac{1}{2}\). Applying \(\ln\) to both sides enables us to isolate \(x^2\), which is a key step in finding the solution.

Square Root

The square root is a fundamental concept in mathematics, often encountered in algebra, geometry, and beyond. Simply put, the square root of a number \(a\), denoted as \(\sqrt{a}\), is a value that, when multiplied by itself, gives \(a\). Therefore, \(x^2 = \ln(2)\) requires us to find \(x\), which involves taking the square root of \(\ln(2)\).

Key properties to remember include:

• \(\sqrt{x^2} = |x|\), meaning it yields the absolute value of \(x\).
• \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\).
• Square roots involve two solutions: the positive and negative root.

In our exercise, solving \(x = \pm \sqrt{\ln(2)}\) reflects this property: there are two solutions, one positive and one negative, representing points on the graph. This is why both \(x = \sqrt{\ln(2)}\) and \(x = -\sqrt{\ln(2)}\) are valid solutions.

Function Graph

A function graph visually represents the relationship between variables, often \(x\) and \(y\). It allows you to see at a glance how changing one variable affects the other. In the equation \(f(x) = 1 - e^{-x^2}\), the graph helps illustrate where the function equals a specific \(y\)-coordinate, like \(\frac{1}{2}\).

Visual Insights:

• The behavior of the graph depends on the nature and type of function.
• Exponential functions like \(e^{-x^2}\) have a distinct curved shape, asymptotically approaching a line.
• The graph of \(f(x) = 1 - e^{-x^2}\) will curve downwards from \(y = 1\) and approach \(y = 0\), never quite reaching it.

Graphing provides a visual verification of the solutions obtained algebraically. According to our solution, there are two \(x\)-values (at \(\pm \sqrt{\ln(2)}\)) for which \(f(x) = \frac{1}{2}\), meaning our function graph intersects \(y = \frac{1}{2}\) at two distinct points.

y-coordinate

The \(y\)-coordinate indicates a point's vertical position on a graph. It's the output value of a function when you input an \(x\)-value. In mathematical notation, it's often referred to as \(f(x)\).

Understanding the \(y\)-coordinate:

• It's a critical aspect for plotting and interpreting graphs.
• Determines where a function intersects horizontal lines, like \(y = 0, 1, \frac{1}{2}\), etc.
• Given a constant \(y\)-value, you can solve for corresponding \(x\)-coordinates.

In our exercise, we are provided a \(y\)-coordinate of \(\frac{1}{2}\) and asked to determine which \(x\)-values yield that output. This involves setting the function equal to \(\frac{1}{2}\) and solving, which led us to find two potential \(x\)-values. Therefore, our graph intersects the line \(y = \frac{1}{2}\) at two distinct points, illustrating this concept beautifully.