Problem 56 - Thomas Calculus

Question

a. Find the local extrema of each function on the given interval, and say where they occur. b. Graph the function and its derivative together. Comment on the behavior of \(f\) in relation to the signs and values of \(f^{\prime}. \) $$ f(x)=-2 x+\tan x, \quad \frac{-\pi}{2}< x <\frac{\pi}{2} $$

Step-by-Step Solution

Answer

Local max: \(x = -\frac{\pi}{4}\); Local min: \(x = \frac{\pi}{4}\).

Step 1: Find the Derivative

First, we need to find the derivative of the function to identify the local extrema. The given function is \( f(x) = -2x + \tan x \). The derivative is found by differentiating each term: \[ f'(x) = -2 + \sec^2{x} \]

Step 2: Set the Derivative Equal to Zero

To find the critical points, we set the derivative equal to zero:\[ -2 + \sec^2{x} = 0 \]Solving for \( x \), we have:\[ \sec^2{x} = 2 \]Taking the reciprocal, we find \( \cos^2{x} = \frac{1}{2} \), which implies \( \cos{x} = \pm\frac{1}{\sqrt{2}} \). Hence, \( x = \pm\frac{\pi}{4} \) within the interval.

Step 3: Determine the nature of extrema

Examine the sign of \( f'(x) \) around the critical points to determine if these points are local maxima or minima. - For \( x < -\frac{\pi}{4} \), \( f'(x) > 0 \), so \( f(x) \) is increasing.- Between \( x = -\frac{\pi}{4} \) and \( x = \frac{\pi}{4} \), \( f'(x) < 0 \), so \( f(x) \) is decreasing.- For \( x > \frac{\pi}{4} \), \( f'(x) > 0 \) again, so \( f(x) \) is increasing.Thus, \( x = -\frac{\pi}{4} \) is a local maximum and \( x = \frac{\pi}{4} \) is a local minimum.

Step 4: Graph both the function and its derivative

Graph \( f(x) = -2x + \tan x \) and its derivative \( f'(x) = -2 + \sec^2{x} \). Observe the behavior:- \( f(x) \) has a local maximum at \( x = -\frac{\pi}{4} \).- \( f(x) \) decreases and reaches a local minimum at \( x = \frac{\pi}{4} \).- The sign changes in \( f'(x) \) confirm these extrema: positive to negative for a max, and negative to positive for a min.

Key Concepts

Understanding Derivatives

Derivatives play a crucial role in calculus as they help us understand the rate at which things change. In simple terms, a derivative gives us the slope of the tangent line to the function at any given point. Think of a car driving down a twisty road. The derivative tells you how quickly you're turning the steering wheel. In mathematical terms, if you have a function \( f(x) \), its derivative is denoted \( f'(x) \).To find the derivative of \( f(x) = -2x + \tan x \), you take the derivative of each part separately:

• The derivative of \(-2x\) is \(-2\).
• The derivative of \( \tan x \) is \( \sec^2{x} \), which is part of trigonometric derivatives.

So, putting these together, we get \( f'(x) = -2 + \sec^2{x} \).Why is this important? Because by analyzing \( f'(x) \), we can determine where the function \( f(x) \) reaches local highs and lows, known as local extrema.

Graphing Functions and Their Derivatives

Graphing functions along with their derivatives provides a visual understanding of how these mathematical concepts relate. When you look at the graph of \( f(x) = -2x + \tan x \) along with its derivative \( f'(x) = -2 + \sec^2{x} \), there are certain patterns to observe.The graph of the function \( f(x) \) shows how it curves and twists through the specified interval. Local extrema appear at critical points, where the derivative \( f'(x) \) equals zero. At these points, the graph of \( f'(x) \) will intersect the x-axis.What do we learn from this? When the graph of \( f'(x) \) is above the x-axis, \( f(x) \) is increasing, and when \( f'(x) \) is below the x-axis, \( f(x) \) is decreasing. Changes in the sign of \( f'(x) \) guide us towards identifying local maxima and minima on the graph of \( f(x) \). This visual relationship helps us understand the behavior of functions far better than just using equations.

Understanding Critical Points

Critical points are where the magic of calculus shines. These are the points on a function where the slope (derivative) is zero or undefined. They are crucial for identifying where a function reaches its tallest peak or deepest valley, which are called local extrema.To find these critical points, we take the derivative \( f'(x) \) and set it equal to zero or check where it's undefined. For our function \( f(x) = -2x + \tan x \), we discovered:

• Solving \(-2 + \sec^2{x} = 0\) gives us \( \sec^2{x} = 2 \).
• This implies \( \cos^2{x} = \frac{1}{2} \), leading to critical points \( x = \pm\frac{\pi}{4} \) within the interval \( \frac{-\pi}{2} < x < \frac{\pi}{2} \).

These points are crucial for understanding where the function changes its behavior. By checking the sign of \( f'(x) \) before and after these points:

• If \( f'(x) \) changes from positive to negative, we have a local maximum.
• If \( f'(x) \) changes from negative to positive, we have a local minimum.

Therefore, knowing how to find and interpret critical points helps us map out the landscape of a function effectively.