Problem 67 - Precalculus with Limits

Question

In Exercises 63-70, use the function and its derivative to determine any points on the graph of \(f\) at which the tangent line is horizontal. Use a graphing utility to verify your results. \(f(x) = x^2 e^x, \quad f'(x) = x^2 e^x + 2xe^x\)

Step-by-Step Solution

Answer

The points on the graph where the tangent line is horizontal are at x = 0 and x = -2.

Step 1:Seperate the common terms

Rewrite the derivative by taking the common terms out. We get \( f'(x) = e^x (x^2 + 2x) \).

Step 2: Find the zeros of the derivative

Set the derivative equal to zero and solve for \( x \), \( e^x (x^2 + 2x) = 0 \). This equation states that either the term \( e^x \) is zero or the term \( x^2 + 2x \) is zero. However, note that \( e^x \) can never be zero for any real number \( x \). Hence, the zeros of the derivative will be the roots of the quadratic equation \( x^2 + 2x = 0 \).

Step 3: Solve the quadratic equation

Solve the quadratic equation \( x^2 + 2x = 0 \) by factoring to find x is 0 or -2.

Step 4: Verify the result

Verify the results using a graphing utility to plot the function and its derivative. The x-values at which the derivative hits the x-axis (i.e., has a value of 0) are the points at which the function will have a horizontal tangent line.

Key Concepts

Understanding Derivatives

The derivative of a function gives us great insight into the rate of change of that function. In simple terms, it shows how a function increases or decreases at any given point. When we talk about tangent lines, the derivative tells us the slope of these lines. If a tangent line is horizontal at a certain point, it means that the slope, or the derivative, is zero at that point.
In the exercise given, the derivative is written as \(f'(x) = x^2 e^x + 2x e^x\). When dealing with derivatives, a common strategy is to factor out terms to simplify the equation. This simplifies our task of finding points where the derivative is zero, which are crucial for locating horizontal tangents.
To find the zero points, the equation is rewritten: \(f'(x) = e^x (x^2 + 2x)\). You set the entire expression to zero, but remember, \(e^x\) is never zero. Therefore, the task is reduced to solving \(x^2 + 2x = 0\). This is a clever calculus trick that makes solving such equations both manageable and insightful.

Quadratic Equations Simplified

Quadratic equations are a type of polynomial equation represented in the standard form \(ax^2 + bx + c = 0\). Solving these equations often involves factoring, using the quadratic formula, or completing the square. In our exercise, the derivative \(x^2 + 2x = 0\) is a quadratic equation.
To solve it, you start by factoring. Firstly, both terms have an \(x\) in common, so pull that out: \(x(x + 2) = 0\). From this, we can deduce two solutions: \(x = 0\) and \(x = -2\).
These solutions indicate the points on the graph where the tangent is horizontal. Understanding these steps in solving quadratic equations is key not just in calculus but also in many mathematical contexts.

Using Graphing Utilities

Graphing utilities are incredibly useful tools for visualizing mathematical functions and their derivatives. They can provide a quick and accurate picture of where functions change, increase, or decrease. In our exercise, after finding the solutions \(x = 0\) and \(x = -2\), a graphing utility helps verify these results.
By inputting the function \(f(x) = x^2 e^x\) and its derivative \(f'(x) = e^x (x^2 + 2x)\) into a graphing calculator or software, you can efficiently observe where the derivative crosses the x-axis. This crossing signifies a zero point, confirming a horizontal tangent.
Graphing utilities do more than just verify solutions. They allow you to explore and understand behaviors of functions you might not easily deduce algebraically. This visual approach offers another dimension of learning, making complex ideas more tangible.