Problem 22 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Find and classify all critical points. Determine whether or not \(f\) attains an absolute maximum and absolute minimum value. If it does, determine the absolute maximum and/or minimum value. \(f(x)=\frac{x}{e^{x}}\)

Step-by-Step Solution

Answer

The critical point of the function \(f(x) = \frac{x}{e^{x}}\) is at \(x = 1\). This point is a maximum. The function attains an absolute maximum value of \(1/e\) at \(x = 1\), and does not have an absolute minimum.

Step 1: Compute the Derivative

Start by finding the derivative of the function \(f(x)\) using the quotient rule:\[f'(x) = \frac{(e^x \cdot 1 - x \cdot e^x)}{(e^x)^2}\]After simplifying, the derivative is :\[f'(x) = \frac{1 - x}{e^x}\]

Step 2: Find Critical Points

To find the critical points, set the derivative equal to zero and solve for \(x\):\[ \frac{1 - x}{e^x} = 0\] Since the denominator is never zero (since \(e^x\) is never zero for all \(x\) ), we simply need to solve the numerator for zero:\[1 - x = 0 \]Therefore, \(x = 1\) is the only critical point.

Step 3: Classify Critical Points

To classify the critical point as maxima, minima, or neither, inspect the derivative's sign to the left and right of \(x = 1\). If we take a number less than \(1\), say \(0.5\), the derivative is positive, as \(\frac{1 - 0.5}{e^{0.5}} > 0\). Taking a number larger than \(1\), say \(1.5\), the derivative becomes negative as \(\frac{1 - 1.5}{e^{1.5}} < 0\). Hence as the function changes from increasing to decreasing at \(x = 1\), it is a maximum

Step 4: Find Absolute Maxima/Minima

The maximum value of the function on the interval \(-\infty < x < \infty\) is the value at the critical point \(x = 1\), which is \(f(1) = 1/e\). Since the function \(f(x)\) tends to zero as \(x\) approaches either \(-\infty\) or \(+\infty\), there is no absolute minimum value.

Key Concepts

Quotient Rule

The Quotient Rule is a technique used to find the derivative of a function that is the ratio of two differentiable functions. When dealing with a function of the form \( f(x) = \frac{u(x)}{v(x)} \), where both \( u(x) \) and \( v(x) \) are differentiable, the derivative \( f'(x) \) can be calculated with the formula:

• \( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \)

It's essential to memorize this formula for ease in tackling problems like that of our exercise where \( f(x) = \frac{x}{e^x} \). Here, the numerator \( u(x) = x \) has a derivative of 1, and the denominator \( v(x) = e^x \) has a derivative of \( e^x \). Following the formula, we find:

• \( f'(x) = \frac{e^x \cdot 1 - x \cdot e^x}{(e^x)^2} \)
• After simplification: \( f'(x) = \frac{1 - x}{e^x} \)

This gives us both the tool and the method to move on to identifying critical points.

Derivative Analysis

Derivative analysis involves using the derivative of a function to determine the behavior of the function. This is crucial for finding critical points. Critical points occur where the derivative is either zero or undefined, highlighting potential maxima, minima, or points of inflection.For the given function \( f(x) = \frac{x}{e^x} \), the derivative \( f'(x) = \frac{1 - x}{e^x} \) was found using the quotient rule. Here, the derivative simplifies to zero when the numerator equals zero because the exponential function \( e^x \) in the denominator is never zero for real numbers. Therefore, we solve:

• \( 1 - x = 0 \)
• This gives us \( x = 1 \) as the only critical point.

Once we locate the critical points, we can use these to interpret the function's behavior by testing intervals around \( x = 1 \). Observing the sign change of \( f'(x) \) at \( x = 1 \) (from positive to negative) confirms a local maximum at this point.

Absolute Extrema

In calculus, absolute extrema refer to the highest and lowest values a function can reach on a given interval. Absolute maximum and minimum values are assessed not just at critical points, but also at endpoints if defined on a closed interval. For the function \( f(x) = \frac{x}{e^x} \), we look to see if the function has an absolute maximum or minimum over \(-\infty < x < \infty\). During our analysis, we saw that the function exhibited its highest point at the critical point \( x = 1 \), where \( f(1) = \frac{1}{e} \). As \( x \) approaches \(-\infty\) or \(+\infty\), the function tends towards zero:

• This is because the exponential growth in the denominator outpaces the linear growth in the numerator.

Thus, while the function has an absolute maximum value of \( \frac{1}{e} \), it does not have an absolute minimum value. This is because it's always decreasing towards, but never actually reaching, zero.