Question

A function \(f(x)\) is given. Find the \(x\) values where \(f^{\prime}(x)\) has a relative maximum or minimum. \(f(x)=x^{2} e^{x}\)

Short answer

The derivative has a relative maximum at \( x = -2 \) and a relative minimum at \( x = 0 \).

Step-by-step solution

Find the First Derivative

To find the critical points where the derivative has a relative maximum or minimum, we first need the derivative of the function. Given \( f(x) = x^2 e^x \), we'll use the product rule. If \( u = x^2 \) and \( v = e^x \), then \( f'(x) = u'v + uv' \). Here, \( u' = 2x \) and \( v' = e^x \). Therefore, \( f'(x) = 2x e^x + x^2 e^x \).

Simplify the Derivative

Factor the derivative \( f'(x) = 2x e^x + x^2 e^x \) to make it easier to handle. Factoring out \( e^x \), we have \( f'(x) = e^x (2x + x^2) = e^x x(2 + x) \).

Find Critical Points of the Derivative

The critical points for relative extrema occur where the derivative equals zero or is undefined. Since \( e^x \) is never zero, set the factored form equal to zero: \( x(2 + x) = 0 \). This gives \( x = 0 \) and \( x = -2 \).

Take the Second Derivative

Find the second derivative, \( f''(x) \), to identify the concavity of \( f'(x) \). Start from the simplified first derivative: \( f'(x) = e^x x(2 + x) \). Apply the product rule and derivative of a product: \( f''(x) = e^x (1)(2 + x) + e^x x(1) \), simplifying to \( f''(x) = e^x (2 + 2x + x) = e^x (2 + 3x) \).

Use the Second Derivative to Identify Extrema

Evaluate the second derivative at the critical points. For \( x = 0 \), \( f''(0) = e^0 (2 + 3(0)) = 2 \), which is positive, indicating \( f'(x) \) has a relative minimum at \( x = 0 \). For \( x = -2 \), \( f''(-2) = e^{-2} (2 + 3(-2)) = -4e^{-2} \), which is negative, indicating \( f'(x) \) has a relative maximum at \( x = -2 \).

Key concepts

Derivative

Derivatives are quite fascinating! At the core, a derivative represents the rate at which a function changes at any given point. Think of it like the function's speedometer, showing how fast it is moving upwards, downwards, or perhaps not moving at all.
To find the derivative of a function, you might use various rules, like the product or chain rule. The derivative is denoted by \( f'(x) \), where \( f(x) \) is the original function. In our exercise, we started with the function \( f(x) = x^2 e^x \) and found its derivative. This involves using the product rule, a crucial tool when you have a function created by multiplying two separate functions.
Understanding derivatives helps you get a grip on critical points and concavity, essential for analyzing the behavior of functions. Derivatives are the stepping stones to discovering how functions behave, giving you insight into changes happening faster than your eye can see.

Critical Points

Critical points are the suspenseful moments in the life of a function. They are points where the derivative \( f'(x) \) is zero or undefined. These points are significant because they can signal a change, potentially marking local maxima, minima, or even points of inflection (where the function changes its concavity).
In the exercise given, to find these critical moments, we set our derived function \( f'(x) = e^x x(2 + x) \) to zero. Solving this indicates where the nuanced dance of change slows down, giving us \( x = 0 \) and \( x = -2 \). At these points, the function may change direction, revealing an extremum of the original function \( f(x) \). Whether these points are turning left or right (concave up or down) is determined by further exploration with second derivatives.
Critical points are the heroes of function analysis, pinpointing where the most interesting action happens.

Concavity

Concavity gives us a deeper understanding of the function's shape and behavior. When a function is concave up, it appears like a cup holding water—imagine a smiley face. A concave down function, by contrast, appears like an upside-down cup, similar to a frown face.
The second derivative \( f''(x) \) is your tool to determine concavity. If \( f''(x) > 0 \), the function is concave up around that point. If \( f''(x) < 0 \), the function is concave down. In our exercise, evaluating \( f''(x) \) at critical points tells us the shape's nature at those moments: a minimum at \( x = 0 \) and a maximum at \( x = -2 \).
Understanding concavity is crucial because it helps predict behavior, ensuring you capture every peak and trough in a function's journey.

Product Rule

The product rule is a lifesaver for dealing with derivatives of products of multiple functions. When you have a function such as \( f(x) \) created by multiplying two separate functions \( u \) and \( v \), the product rule shows you how to derive it correctly.
Mathematically, the rule states: if \( f(x) = u(x) v(x) \), then \( f'(x) = u'(x)v(x) + u(x)v'(x) \). This implies taking derivatives of each function individually, then combining them with the original functions in a specific manner. In our case, \( f(x) = x^2 e^x \), we set \( u = x^2 \) and \( v = e^x \), using the product rule to reveal the magic behind the change rate of the entire function.
Mastering the product rule enriches your calculus toolkit, empowering you to tackle complex functions with confidence.