Problem 36 - Calculus Single Variable

Question

Calculate the first and second derivatives of the given expression, and classify its local extrema. \(x 2^{x}\)

Step-by-Step Solution

Answer

The function has a local minimum at \(x = -\frac{1}{\ln(2)}\).

Step 1: Differentiate the Expression

The function given is \(f(x) = x \cdot 2^x\). We need to find its first derivative. Use the product rule, which states that \((uv)' = u'v + uv'\). Let \(u = x\) and \(v = 2^x\). 1. Derivative of \(u\) is \(u' = 1\).2. Derivative of \(v\) involves the base of the exponential function, giving us \(v' = 2^x \ln(2)\).Thus, the first derivative is:\[ f'(x) = 1 \cdot 2^x + x \cdot 2^x \ln(2) = 2^x (1 + x \ln(2)) \]

Step 2: Differentiate a Second Time

Now find the second derivative by differentiating \(f'(x) = 2^x (1 + x \ln(2))\) again.First, note this is also a product and use the product rule again:Let \(u = 2^x\) and \(v = 1 + x \ln(2)\).1. Derivative of \(u\) is \(u' = 2^x \ln(2)\).2. Derivative of \(v\) is \(v' = \ln(2)\).Using the product rule, the second derivative is:\[ f''(x) = (2^x \ln(2))(1 + x \ln(2)) + 2^x(\ln(2)) = 2^x \ln(2)(1 + x \ln(2) + 1) = 2^x \ln(2)(2 + x \ln(2)) \]

Step 3: Determine Critical Points

Find critical points by setting the first derivative to zero:\[ f'(x) = 2^x (1 + x \ln(2)) = 0 \]Solving this equation gives:1. \(2^x eq 0\), so set \(1 + x \ln(2) = 0\).2. Solve for \(x\):\[ x \ln(2) = -1 \Rightarrow x = -\frac{1}{\ln(2)} \]

Step 4: Classify Local Extrema

With the critical point \(x = -\frac{1}{\ln(2)}\), use the second derivative test:Evaluate \(f''\left(-\frac{1}{\ln(2)}\right)\):1. Substitute \(x = -\frac{1}{\ln(2)}\) into \(f''(x) = 2^x \ln(2)(2 + x \ln(2))\).2. Since \(f''(x) > 0\) for all \(x\), the second derivative is positive.Therefore, \(x = -\frac{1}{\ln(2)}\) corresponds to a local minimum.

Key Concepts

Product Rule

The product rule is an essential tool in calculus, especially when differentiating functions that are the product of two other functions. It helps us find the derivative of products correctly. Here's a simple way to remember it: if you have a function that is the product of two functions, say \( u(x) \) and \( v(x) \), the derivative is given by

• \((uv)' = u'v + uv'\).

This means we take the derivative of the first function and multiply it by the second, and add it to the product of the first function and the derivative of the second.
In our exercise, we apply this rule to the function \(f(x) = x \cdot 2^x\). Here, we identified \(u = x\) and \(v = 2^x\).

• The derivative of \(u\), which is \(x\), is simply 1.
• The derivative of \(v\), which involves the exponential \(2^x\), requires us to consider the natural logarithm of the base, resulting in \(v' = 2^x \ln(2)\).

Combining these, we get the first derivative as \(f'(x) = 2^x (1 + x \ln(2))\).
The product rule is crucial in ensuring all components are accounted for, especially with functions involving both polynomials and exponentials.

Second Derivative

The second derivative gives us insight into the concavity of the function. It helps in determining where the function changes direction, which is crucial in finding points of inflection and local extrema. To find the second derivative, we differentiate the first derivative. This often requires applying the product rule again if the first derivative is still a product of functions.
In the exercise, after obtaining \(f'(x) = 2^x (1 + x \ln(2))\), we applied the product rule again:

• Let \(u = 2^x\) with derivative \(u' = 2^x \ln(2)\),
• and \(v = 1 + x \ln(2)\) with derivative \(v' = \ln(2)\).

Using the product rule, the second derivative is:

• \(f''(x) = 2^x \ln(2)(2 + x \ln(2))\).

This expression is key when using the second derivative test to classify critical points, letting us understand the behavior of the function's curvature.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points are important because they are the locations where the function could potentially have a local maximum or minimum. To find critical points, we solve \(f'(x) = 0\).
In our problem, we found the critical points by setting the first derivative equal to zero:

• \(f'(x) = 2^x (1 + x \ln(2)) = 0\).

Since \(2^x\) is never zero, we only need to solve:

• \(1 + x \ln(2) = 0\).

Solving this equation gives us:

• \(x = -\frac{1}{\ln(2)}\).

This process identifies where the function's slope is zero or undefined, pointing us to potential extrema or inflections in the graph.

Local Extrema

Local extrema refer to local maximum or minimum points on the graph of a function. To determine whether a critical point is a local maximum or minimum, we use the second derivative test. If the second derivative is positive at a critical point, the function is concave up and the point is a local minimum. If it's negative, the function is concave down and the point is a local maximum.
For the exercise, we evaluate the second derivative at the critical point \(x = -\frac{1}{\ln(2)}\) by substituting back into:

• \(f''(x) = 2^x \ln(2)(2 + x \ln(2))\).

Finding \(f''(x) > 0\) indicates the function is concave up at this point. Thus, \(x = -\frac{1}{\ln(2)}\) corresponds to a local minimum.
Understanding local extrema is essential for sketching graphs and solving optimization problems, as it tells us where functions reach their highest or lowest values locally.