Problem 15 - Calculus for Biology and Medicine

Question

Find \(c\) so that \(f^{\prime}(c)=0 .\) \(f(x)=\sin \left(\frac{\pi}{2} x\right)\)

Step-by-Step Solution

Answer

The value of \( c \) is 1.

Step 1: Find the Derivative of f(x)

Given the function \( f(x) = \sin \left( \frac{\pi}{2} x \right) \), we need to find its derivative. Using the chain rule, the derivative of \( \sin(ax) \) is \( a \cos(ax) \), thus the derivative is \( f'(x) = \frac{\pi}{2} \cos \left( \frac{\pi}{2} x \right) \).

Step 2: Set the Derivative to Zero

Set the derivative \( f'(x) = \frac{\pi}{2} \cos \left( \frac{\pi}{2} x \right) \) equal to zero to find the critical points: \( \frac{\pi}{2} \cos \left( \frac{\pi}{2} x \right) = 0 \). This simplifies to \( \cos \left( \frac{\pi}{2} x \right) = 0 \).

Step 3: Solve the Equation for x

The cosine function is zero at odd multiples of \( \frac{\pi}{2} \). So \( \frac{\pi}{2} x = \frac{\pi}{2} + n\pi \), where \( n \) is an integer. Solving for \( x \), we get \( x = 1 + 2n \) or \( x = 1, 3, 5, \ldots \).

Step 4: Conclusion on the Value of c

Since the problem asks for any \( c \) where \( f'(c) = 0 \), we can choose \( c = 1 \) as a representative solution, although \( c = 1 + 2n \) with \( n \) as an integer is the general solution.

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of composite functions. A composite function is when one function is nested inside another. To better understand this, let's consider a function like \( f(x) = \sin\left(\frac{\pi}{2} x\right) \). This can be viewed as a composition of two functions: an outer function \( g(u) = \sin(u) \) and an inner function \( u = \frac{\pi}{2} x \).

• The chain rule states that the derivative of a composite function \( f(x) = g(u) \) where \( u = h(x) \) is found as \( f'(x) = g'(h(x)) \cdot h'(x) \).
• In our example, the derivative of the outer function \( \sin(u) \) is \( \cos(u) \), and the derivative of the inner function \( \frac{\pi}{2} x \) with respect to \( x \) is \( \frac{\pi}{2} \).
• Therefore, the derivative of \( f(x) = \sin\left(\frac{\pi}{2} x\right) \) is \( f'(x) = \cos \left(\frac{\pi}{2} x\right) \cdot \frac{\pi}{2} = \frac{\pi}{2} \cos \left(\frac{\pi}{2} x\right) \).

Mastering the chain rule allows you to break down and differentiate complex compositions effectively.

Critical Points

Critical points of a function occur where its derivative is zero or undefined. These points can indicate where the function reaches local maxima, minima, or saddle points. To find these points:

• First, compute the derivative. For our function \( f(x) = \sin\left(\frac{\pi}{2} x\right) \), the derivative is \( f'(x) = \frac{\pi}{2} \cos \left(\frac{\pi}{2} x \right) \).
• Set this derivative equal to zero: \( \frac{\pi}{2} \cos\left(\frac{\pi}{2} x\right) = 0 \).
• Simplifying, it reduces to \( \cos\left(\frac{\pi}{2} x\right) = 0 \).
• This implies critical points are at values of \( x \) making the cosine function zero, which are the odd multiples of \( \frac{\pi}{2} \).

Identifying critical points is crucial as it provides insight into the behavior of a function across its domain.

Solving Trigonometric Equations

Trigonometric equations can often look complex due to the wave-like behavior of sine, cosine, and other trigonometric functions. Solving these equations requires understanding the properties of these periodic functions:

• For our equation \( \cos\left(\frac{\pi}{2} x\right) = 0 \), we solve by identifying that the cosine function is zero at angles such as \( \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots \).
• Setting \( \frac{\pi}{2} x = \frac{\pi}{2} + n\pi \), where \( n \) is an integer, encapsulates the general solution for this equation.
• Solving for \( x \) gives \( x = 1 + 2n \), meaning solutions are integers \( 1, 3, 5, \ldots \).

Understanding how to navigate these trigonometric equations by leveraging their identities and periodic properties can simplify finding solutions significantly.