Question

A function \(f(x)\) is given. Find the critical points of \(f\) and use the Second Derivative Test, when possible, to determine the relative extrema. \(f(x)=x^{2} e^{x}\)

Short answer

Critical points are at \(x = 0\) (minimum) and \(x = -2\) (maximum).

Step-by-step solution

Find the first derivative

To find the critical points of the function, we first need the derivative, which is the first step. Using the product rule for differentiation on \(f(x) = x^2 e^x\), we get \(f'(x) = 2x e^x + x^2 e^x\).

Factor the first derivative

Next, we factor the first derivative to simplify finding the critical points. \(f'(x) = x e^x (2 + x)\).

Find critical points

Set the derivative equal to zero to find the critical points: \(x e^x (2 + x) = 0\). This gives critical points at \(x = 0\) and \(x = -2\).

Find the second derivative

Now, we find the second derivative to use the Second Derivative Test. Differentiating \(f'(x) = 2x e^x + x^2 e^x\) again, using the product rule for each term, we get \(f''(x) = 2e^x + 4x e^x + x^2 e^x = e^x (2 + 4x + x^2)\).

Apply the Second Derivative Test

We use the second derivative to test each critical point. For \(x = 0\), \(f''(0) = e^0 (2 + 4(0) + 0^2) = 2\), which is positive, indicating a relative minimum. For \(x = -2\), \(f''(-2) = e^{-2} (2 + 4(-2) + (-2)^2) = e^{-2} \cdot (-2) = -\frac{2}{e^{2}}\), which is negative, indicating a relative maximum.

Key concepts

Critical Points

Critical points are special points on the graph of a function where the derivative is zero or undefined. These points help us determine where the function might have a relative maximum, minimum, or a point of inflection. To find the critical points of a function:

• Compute the first derivative of the function.
• Set the derivative equal to zero and solve for the variable.
• Identify any points where the derivative does not exist.

In the exercise, the first derivative was determined as \(f'(x) = x e^x (2 + x)\). Solving \(x e^x (2 + x) = 0\) gives the critical points \(x = 0\) and \(x = -2\). These points are essential in determining the behavior of the function at those specific x-values.

Product Rule

The product rule is a formula used to find the derivative of the product of two functions. It states that if you have two functions, \(u(x)\) and \(v(x)\), the derivative of their product \(u(x)v(x)\) is given by:

• \((uv)' = u'v + uv'\)

This rule is crucial when dealing with expressions where two differentiable functions are multiplied together. In the given function \(f(x) = x^2 e^x\), both \(x^2\) and \(e^x\) are differentiable, and hence we apply the product rule:

• The derivative \(u' = 2x\) when \(u = x^2\),
• The derivative \(v' = e^x\) when \(v = e^x\).

By using the product rule, we obtained the first derivative \(f'(x) = 2x e^x + x^2 e^x\), which was then factorized for further analysis.

Relative Extrema

Relative extrema refer to the relative maximum or minimum values of a function within a certain interval. These points are critical in understanding the overall shape and turning points on a graph. To determine relative extrema, we often use:

• The First Derivative Test.
• The Second Derivative Test.

In this case, the Second Derivative Test was employed. The critical point \(x = 0\) was found to have a positive second derivative, \(f''(0)=2\), indicating a relative minimum. Meanwhile, at \(x = -2\), the second derivative is \(-\frac{2}{e^2}\), which is negative, suggesting a relative maximum. Identifying these points allows us to conclude how the function behaves locally around \(x = 0\) and \(x = -2\).

First Derivative

The first derivative of a function represents the rate of change or the slope of the function at any given point. It is pivotal for identifying critical points and understanding the behavior of a function. The steps to find the first derivative include:

• Apply the rules of differentiation, such as the product rule, power rule, and chain rule where necessary.
• Simplify the resultant expression if possible.
• Use the first derivative to find critical points by setting it equal to zero.

In our example, the first derivative is \(f'(x) = 2x e^x + x^2 e^x\), which through simplification is expressed as \(f'(x) = x e^x (2 + x)\). The first derivative is not just crucial for finding critical points, but it also paves the way for using the second derivative to confirm and classify those points.