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Magazine Chapter 4: Problem 47
Use any method to find the relative extrema of the function \(f .\) $$ f(x)=e^{2 x}-e^{x} $$
Short Answer
Expert verified
The function has a relative minimum of \(-\frac{1}{4}\) at \( x = \ln(\frac{1}{2}) \).
Step by step solution
01
Determine the First Derivative
To find the relative extrema of a function, we start by finding its first derivative. The function given is \( f(x) = e^{2x} - e^x \). We apply the derivative rules for exponential functions:\( f'(x) = \frac{d}{dx} (e^{2x}) - \frac{d}{dx} (e^x) = 2e^{2x} - e^x \).
02
Find Critical Points
Critical points occur where the first derivative is zero or undefined. In this case, we solve for \( f'(x) = 0 \):\[ 2e^{2x} - e^x = 0 \]We can factor out \( e^x \) from the equation:\[ e^x (2e^x - 1) = 0 \]Setting each factor equal to zero gives:1. \( e^x = 0 \) (which has no real solutions since \( e^x > 0 \) for all real \( x \))2. \( 2e^x - 1 = 0 \) which simplifies to \( e^x = \frac{1}{2} \).Therefore, \( x = \ln(\frac{1}{2}) \).
03
Second Derivative Test
To determine if the critical point is a relative minimum or maximum, we use the second derivative test. We find the second derivative of the function:\( f''(x) = \frac{d}{dx} (2e^{2x} - e^x) = 4e^{2x} - e^x \).Substitute \( x = \ln(\frac{1}{2}) \) into the second derivative:\[ f''(\ln(\frac{1}{2})) = 4e^{2\ln(\frac{1}{2})} - e^{\ln(\frac{1}{2})} = 4\left(\frac{1}{2}\right)^2 - \frac{1}{2} \]\[ = 4 \times \frac{1}{4} - \frac{1}{2} = 1 - \frac{1}{2} = \frac{1}{2} \]Since \( f''(\ln(\frac{1}{2})) > 0 \), the function has a relative minimum at \( x = \ln(\frac{1}{2}) \).
04
Find the Relative Minimum Value
To find the value of the relative minimum, substitute \( x = \ln(\frac{1}{2}) \) back into the original function:\[ f(\ln(\frac{1}{2})) = e^{2\ln(\frac{1}{2})} - e^{\ln(\frac{1}{2})} \]\[ = \left(\frac{1}{2}\right)^2 - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \]So, the relative minimum value of the function is \(-\frac{1}{4}\) at \( x = \ln(\frac{1}{2}) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Relative Extrema
In calculus, relative extrema refer to the highest or lowest points within a given interval on a continuous curve. These points can be identified as relative maxima or minima. When you are asked to find the relative extrema of a function, you are essentially asked to find where the function has peaks (maximum) or troughs (minimum) compared to the nearby points in the graph.
To find these points, start by finding the first derivative of the function. This derivative tells you how the function is behaving - whether it's increasing or decreasing at any given point. Particularly, relative extrema occur at critical points which are detailed further below. Use the first derivative test or the second derivative test (as explained in later sections) to determine if these points are indeed maxima or minima.
To find these points, start by finding the first derivative of the function. This derivative tells you how the function is behaving - whether it's increasing or decreasing at any given point. Particularly, relative extrema occur at critical points which are detailed further below. Use the first derivative test or the second derivative test (as explained in later sections) to determine if these points are indeed maxima or minima.
Critical Points
Critical points are special points where the function's slope is zero or undefined, indicating potential relative maxima, minima, or inflection points. It's where the function changes its increasing or decreasing nature. Identifying these points is the first step in the hunt for relative extrema.
To locate them, calculate the first derivative of the function and solve for when this derivative is equal to zero. For the function given, the critical point found was at \(x = \ln(\frac{1}{2}) \).
Importantly, not all critical points are extrema; further tests are necessary to determine their nature. This leads us to the next concept: the second derivative test.
To locate them, calculate the first derivative of the function and solve for when this derivative is equal to zero. For the function given, the critical point found was at \(x = \ln(\frac{1}{2}) \).
Importantly, not all critical points are extrema; further tests are necessary to determine their nature. This leads us to the next concept: the second derivative test.
Second Derivative Test
The second derivative test is a convenient way to classify the nature of critical points. After finding the second derivative of the function, substitute the critical points into this second derivative. This will help determine convexity or concavity of the function at those points.
For the given exercise, using the second derivative test, the critical point \(x = \ln(\frac{1}{2})\) was verified to be a relative minimum, as the second derivative was positive at this point.
- If the second derivative at the critical point is positive \(f''(x) > 0\), the function is concave up, and thus the point is a relative minimum.
- If \(f''(x) < 0\), the function is concave down, indicating a relative maximum.
For the given exercise, using the second derivative test, the critical point \(x = \ln(\frac{1}{2})\) was verified to be a relative minimum, as the second derivative was positive at this point.
Exponential Functions
Exponential functions are functions of the form \(e^{x}\) or similar, where \(e\) is the base of the natural logarithm, approximately 2.718. These functions grow rapidly, making them particularly important in calculus. Their derivatives and integrals offer unique properties and are frequently encountered in applications across mathematics and sciences.
When differentiating an exponential function like \(e^{ax}\), remember to apply the chain rule. For example, \( \frac{d}{dx}(e^{2x}) = 2e^{2x} \) by bringing down the coefficient of \(x\) as a factor.
Keep in mind that exponential functions never reach zero; they approach it asymptotically. Therefore, equations like \(e^{x} = 0\) will have no real solutions. Understanding these and related properties can greatly help in comprehending how these functions behave and evolve.
When differentiating an exponential function like \(e^{ax}\), remember to apply the chain rule. For example, \( \frac{d}{dx}(e^{2x}) = 2e^{2x} \) by bringing down the coefficient of \(x\) as a factor.
Keep in mind that exponential functions never reach zero; they approach it asymptotically. Therefore, equations like \(e^{x} = 0\) will have no real solutions. Understanding these and related properties can greatly help in comprehending how these functions behave and evolve.
