Problem 37 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the absolute maximum value and the absolute minimum value, if any, of each function. $$ g(x)=(2 x-1) e^{-x} \text { on }[0,4] $$

Step-by-Step Solution

Answer

The absolute maximum value of the function \(g(x) = (2x - 1)e^{-x}\) on the interval [0, 4] is approximately 0.1839 at \(x = \frac{3}{2}\), and the absolute minimum value is -1 at \(x = 0\).

Step 1: Find the first derivative and critical points

Let's find the first derivative of the function \(g(x) = (2x - 1)e^{-x}\) using the product rule. Given: \(u(x) = 2x - 1\) and \(v(x) = e^{-x}\) Find: \(u'(x) = 2\) (derivative of \(2x - 1\)) Find: \(v'(x) = -e^{-x}\) (derivative of \(e^{-x}\)) Now apply the product rule: \(g'(x) = u'(x)v(x) + u(x)v'(x)\) We get: \(g'(x) = 2e^{-x} + (2x - 1)(-e^{-x})\) Next, find the critical points where the derivative equals 0 or is undefined: Critical points: \(g'(x) = 0 \) \(2e^{-x} + (2x - 1)(-e^{-x}) = 0\) Now we must solve for x.

Step 2: Solve for critical points

The equation for the critical points is \(2e^{-x} + (2x - 1)(-e^{-x}) = 0\). We can factor out an \(e^{-x}\) from the equation: \(e^{-x}(2 - (2x-1)) = 0\) This simplifies to: \(e^{-x}(3-2x)=0\) Since the exponential function \(e^{-x}\) is always positive, the equation has a solution if and only if the other factor in the product is equal to zero: \(3 - 2x = 0\) Solving for x, we get: \(x = \frac{3}{2}\) So our only critical point is \(x = \frac{3}{2}\).

Step 3: Evaluate g(x) at critical points and endpoints

Evaluate the given function at the critical point and the endpoints of the interval [0, 4]: - \(g(0) = (2(0) - 1)e^{-0} = -1\) - \(g(\frac{3}{2}) = (2(\frac{3}{2}) - 1)e^{-\frac{3}{2}} \approx 0.1839\) - \(g(4) = (2(4) - 1)e^{-4} \approx -0.0183\)

Step 4: Determine the absolute maximum and minimum values

Comparing the function values, we can clearly see the absolute maximum and minimum: - Absolute maximum value: \(g(\frac{3}{2}) \approx 0.1839\) - Absolute minimum value: \(g(0) = -1\) Thus, the absolute maximum value is approximately 0.1839, and the absolute minimum value is -1.

Key Concepts

Critical Points

Critical points in calculus are values of the variable where the derivative of a function is zero or undefined. These points are significant because they can indicate potential locations for local maxima or minima. To find them, we take the first derivative of the function.The first derivative tells us about the slope at any point on the function's curve:- If the derivative is positive, the function is increasing.- If the derivative is negative, the function is decreasing.- A derivative of zero might indicate a local maximum, a local minimum, or a saddle point.In our exercise we have a function:\[ g(x) = (2x-1)e^{-x} \]Using the product rule, we found the derivative and set it to zero to find critical points:\[ 2e^{-x} + (2x - 1)(-e^{-x}) = 0 \]Solving this, we obtained the critical point at:\[ x = \frac{3}{2} \]This point is crucial in determining where the function might reach maximum or minimum values.

Absolute Maximum

The absolute maximum of a function is the highest value that the function takes on a given interval. It can occur at critical points or at the endpoints of the interval.To determine the absolute maximum, we evaluate the function at the critical points and at the endpoints of the interval we’re interested in. In our example, the interval is from 0 to 4, and our critical point is at \( x = \frac{3}{2} \).We then compute:- \( g(0) = -1 \)- \( g(\frac{3}{2}) \approx 0.1839 \)- \( g(4) \approx -0.0183 \)Out of these values, the largest one is approximately 0.1839 at \( x = \frac{3}{2} \). This means the absolute maximum of \( g(x) \) on the interval \([0,4]\) is approximately 0.1839, occurring at the critical point.

Absolute Minimum

The absolute minimum, on the other hand, is the smallest value that a function can take on a specific interval. Similar to the absolute maximum, it can be at a critical point or at one of the interval's endpoints.We were given the same interval \([0,4]\) to evaluate our function \( g(x) \). After calculating the function values at the critical point and the endpoints, we compare them to decide which is the least.- At \( x = 0 \), we have a function value of \( g(0) = -1 \).- At \( x = \frac{3}{2} \), the value is approximately 0.1839.- At \( x = 4 \), the value is approximately -0.0183.Among these, the smallest value is \( g(0) = -1 \). Hence, the absolute minimum of the function in the given interval \([0,4]\) is \(-1\), occurring at the left endpoint \( x = 0 \).