Question

Locate the critical points of the following functions. Then use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima. $$f(x)=x^2{e}^{-x}$$

Short answer

Answer: The critical points are $x=0$ and $x=2$. The point $x=0$ corresponds to a local minimum, while $x=2$ corresponds to a local maximum.

Step-by-step solution

Calculate the first derivative

To determine the critical points of the given function, we need to find its first derivative. To do this, we'll apply the product rule, since it is a product of two functions, $x^2$ and $e^{-x}$. The product rule states that for any two functions $u(x)$ and $v(x)$, their derivative is given by: $$\frac{d(uv)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$$ Let $u(x)=x^2$ and $v(x)=e^{-x}$. So, $$\frac{du}{dx}=2x,\frac{dv}{dx}=-e^{-x}$$ Using the product rule: $$\frac{df}{dx}=x^2(-e^{-x})+e^{-x}\left(2x\right)=-x^2{e}^{-x}+2x{e}^{-x}$$

Find the critical points

To find the critical points, we need to set the first derivative equal to zero and solve for x: $$-x^2{e}^{-x}+2x{e}^{-x}=0$$ Factor out $x{e}^{-x}$: $$x{e}^{-x}(-x+2)=0$$ So, we get two critical points: $$x=0\text{ or }x=2$$

Calculate the second derivative

To use the second derivative test, we need to find the second derivative of the function. We will do this by differentiating the first derivative that we found in Step 1: $$\frac{d^{2}f}{d{x}^2}=\frac{d}{dx}(-x^2{e}^{-x}+2x{e}^{-x})$$ Use the product rule again and sum the results: $$\frac{d^{2}f}{d{x}^2}=(2x-2)e^{-x}+(-x^2+4x-2)e^{-x}=(x^2-4x+2)e^{-x}$$

Use the Second Derivative Test

Let's now use the second derivative test to determine whether the critical points correspond to local maxima or local minima. To do this, evaluate the second derivative at each critical point. If the value is greater than zero, then it's a local minimum. If the value is less than zero, then it's a local maximum. If it's equal to zero, then the test is inconclusive. For $x=0:$ $$\frac{d^{2}f}{d{x}^2}{|}_{x=0}=(0^2-4\cdot 0+2)e^0=2>0$$ Thus, $x=0$ corresponds to a local minimum. For $x=2:$ $$\frac{d^{2}f}{d{x}^2}{|}_{x=2}=(2^2-4\cdot 2+2)e^{-2}=-4e^{-2}<0$$ Thus, $x=2$ corresponds to a local maximum. The critical points $x=0$ and $x=2$ correspond to a local minimum and a local maximum, respectively.

Key concepts

Critical Points

Critical points are places on a graph where the slope of the function is zero or undefined. To find these points, you need to determine where the first derivative of the function equals zero. These points are important because they potentially mark where the function reaches a local maximum or minimum.
In our example, the function is given as $f(x)=x^2{e}^{-x}$. First, we find the first derivative. Apply the product rule on the terms, set the derivative to zero, and solve for $x$. This process reveals our critical points: $x=0$ and $x=2$. Identifying these points is the first step before using further tests to determine the nature of these points.

Second Derivative Test

The Second Derivative Test helps to identify if our critical points correspond to local maximum or minimum values. This test involves the second derivative of the function.
Once you have the second derivative, evaluate it at each critical point:

• If the second derivative is positive, it suggests the point is a local minimum.
• If it is negative, it suggests a local maximum.
• If the second derivative equals zero, the test does not provide any information.

In our problem, the second derivative evaluated at $x=0$ is positive, so $x=0$ is a local minimum. At $x=2$, the second derivative is negative, indicating a local maximum.

Product Rule

The product rule is a technique used in calculus to find the derivative of a function that is the product of two other functions. For two functions $u(x)$ and $v(x)$, the product rule states: $$\frac{d(uv)}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$$In the provided function $f(x)=x^2{e}^{-x}$, set $u(x)=x^2$ and $v(x)=e^{-x}$. Differentiating each part and applying the product rule helps in finding the derivatives we need for the critical points and second derivative, essential for the remaining tests.

Local Maxima

Local maxima are points where the function value is higher than nearby points. You can use the first and second derivatives to find these.
For instance, if you find $x=2$ to be a critical point and the second derivative at this point is negative, as in our exercise, then we have identified a local maximum. This means the function $f(x)$ increases up to $x=2$ and then decreases, making $x=2$ a "peak" on the graph of the function.

Local Minima

Local minima are points where the function value is less than the nearby points. The second derivative test helps decide these points by showing whether the slope changes from negative to positive at a critical point.
From our example, $x=0$ is identified as a local minimum. At this critical point, the second derivative is positive, confirming it is a point where the function has a "valley," or a local low point, before it starts to rise again.