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Magazine Chapter 27: Problem 22
Solve the given problems by finding the appropriate derivative.
By Newton's method, find the value of \(x\) for which \(y=e^{\cos x}\) is minimum
for \(0
Short Answer
Expert verified
The minimum occurs at
\( x = \pi \).
Step by step solution
01
Find the derivative
First, we need to find the derivative of the function \[ y = e^{\cos x}. \]We use the chain rule for derivatives here. The derivative of \( e^u \) is \( e^u \cdot u' \), where \( u = \cos x \). The derivative of \( \cos x \) is \( -\sin x \). Thus, the derivative is \[-e^{\cos x} \sin x.\]
02
Set the derivative to zero
To find the critical points where the function could reach a minimum, we set the derivative to zero:\[-e^{\cos x} \sin x = 0.\]Since \( e^{\cos x} eq 0 \), the equation simplifies to \( \sin x = 0 \).
03
Solve for critical points
Solve \( \sin x = 0 \) within the interval \( 0 < x < 2\pi \). The solutions are \( x = \pi \). These are the possible points where the minimum could occur.
04
Verify with the second derivative
To ensure that the point we found is indeed a minimum, we check the second derivative. The first derivative is \( -e^{\cos x} \sin x \). Further differentiate to find the second derivative:\[\frac{d}{dx}(-e^{\cos x} \sin x) = e^{\cos x} \sin^2 x - e^{\cos x} \cos x.\]Evaluate this second derivative at \( x = \pi \):\[e^{-1} \cdot 0^2 + e^{-1} \cdot 1 = e^{-1} > 0.\]This indicates that the function has a local minimum at \( x = \pi \).
05
Confirmation using graphing
To confirm our finding, use a graphing calculator to plot \( y = e^{\cos x} \) over the interval \( 0 < x < 2\pi \). Observe the graph to see if there is a local minimum corresponding to our calculated point \( x = \pi \). The graph should show a minimum at this value.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Newton's Method
One powerful technique in calculus for finding roots, or zeros, of a function is Newton's Method. It's iterative and begins with an initial guess. The idea is to use the derivative of a function to approximate its zeroes. To briefly summarize how this works:
- Start with an initial guess, say, \( x_0 \).
- Use the formula \( x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} \) to find a better approximation \( x_{n+1} \).
Chain Rule
When it comes to differentiating complex functions, the Chain Rule is indispensable in calculus. It helps find the derivative of composite functions. If you have a function that is composed of two other functions, say \( y = f(g(x)) \), the Chain Rule states that the derivative is:\[ y' = f'(g(x)) \, g'(x) \]In our problem, we had the function \( y = e^{\cos x} \). Here, consider \( u = \cos x \), and thus, the function becomes \( e^u \). By the Chain Rule, the derivative of \( e^u \) is \( e^u \cdot (-\sin x) \). This gives us a clear, methodical way to handle differentiation when functions nest within each other, making it a must-know for calculus students.
Critical Points
Critical points in a function are where potential local maxima or minima reside, crucial for understanding the behavior of curves. A critical point occurs where the first derivative, \( f'(x) \), equals zero or is undefined. These points suggest locations where the function might change direction.
- First, find the derivative of the function.
- Set this derivative to zero and solve for \( x \).
Graphing Calculator
In today's digital era, graphing calculators are indispensable tools for visualizing and confirming mathematical solutions. To graph a function like \( y = e^{\cos x} \), a graphing calculator can offer a detailed visual representation.
- Input the function into the calculator.
- Set the viewing window to the desired interval, here \( 0 < x < 2\pi \).
- Zoom in on areas of interest to inspect features like minima or maxima.
