Problem 52 - Essential Calculus

Question

Consider the function on the interval \((0,2 \pi)\) For each function, (a) find the open interval(s) on which the function is increasing or decreasing, (b) apply the First Derivative Test to identify all relative extrema, and (c) use a graphing utility to confirm your results. $$ f(x)=\sqrt{3} \sin x+\cos x $$

Step-by-Step Solution

Answer

The function \(f(x) = \sqrt{3} \sin x + \cos x\) is increasing on the intervals \((0, \pi/3)\) and \((4\pi/3, 2\pi)\), while it is decreasing on the interval \((\pi/3, 4\pi/3)\). There is a local maximum at \(x = \pi / 3\) and a local minimum at \( x = 4\pi / 3\).

Step 1: Find the derivative of the function

For the given function \(f(x) = \sqrt{3} \sin x + \cos x\), determine the derivative \(f'(x)\) which is the rate of change of the function at any given point x. Using the formula \(\cos'(x) = -\sin x\) and \(\sin'(x) = \cos x\), we have \(f'(x) = \sqrt{3} \cos x - \sin x\).

Step 2: Determine the intervals where the derivative is positive or negative

Set the derivative to zero to find the critical points. So, we solve for \(x\) in the equation \(\sqrt{3} \cos x - \sin x = 0\) . Using the trigonometric identity, \(\tan x = \sqrt{3}\), we can find that \(x = \pi / 3, 4\pi / 3\), within the interval \((0, 2\pi)\). Then, determine the sign of \(f'(x)\) in the intervals between these critical points to denote where the function is increasing or decreasing.

Step 3: Apply the First Derivative Test to identify relative extrema

On changing from a positive to a negative derivative, the function has a local maximum. In reverse, from negative to positive, it reaches a local minimum. Hence a local maximum is present at \(x = \pi / 3\) and a local minimum at \( x = 4\pi / 3\) based on the signs of the derivative in the intervals we've already calculated.

Step 4: Confirm your results with a graphing utility

Using any available graphing utility (like a graphing calculator or software like Desmos, GeoGebra, etc.), graph the function \(f(x) = \sqrt{3} \sin x + \cos x\). This should visually confirm the intervals of increasing and decreasing you've determined, as well as any local minima or maxima at the points we've found.

Key Concepts

First Derivative Test

The First Derivative Test is a crucial tool in calculus used to locate the relative extrema (which are the local maxima and minima) of a function. To begin with, one must find the derivative of the function, which provides information on its slope or rate of change at any given point. For instance, in our exercise, the function is

\( f(x) = \frac{1}{4}x^4 - x^3 \)

Its derivative, \( f'(x) = x^3 - 3x^2 \) represents how \( f(x) \) changes with respect to \( x \) . This derivative is then set to zero to find the critical points, which are potential locations of extrema.

After identifying the critical points, we investigate the sign of the derivative on either side of each point. If \( f'(x) \) changes from positive to negative at a critical point, it indicates a local maximum. Conversely, if it shifts from negative to positive, we have a local minimum. In our exercise, these changes in sign signify that we have a local maximum at \( x = \frac{1}{4} \) and a local minimum at \( x = 3 \). Using the First Derivative Test allows us to conclude that the function \( f(x) \) increases up to \( x = \frac{1}{4} \) and then decreases, re-increasing after \( x = 3 \) .

Increasing and Decreasing Intervals

Understanding intervals of increase and decrease in a function helps in sketching its graph and predicting its behavior. To determine these intervals for a continuous function, one uses the sign of the first derivative. With our example function,

\( f(x) = \frac{1}{3}x^3 - \frac{1}{2}x^2 \)

we analyze where its derivative \( f'(x) \) is positive (indicating increase) and where it's negative (indicating decrease).

By finding the critical points, where \( f'(x) = 0 \) or is undefined, we divide the domain into intervals. We then test each interval by inserting any number from it into the derivative. If \( f'(x) > 0 \) on an interval, the function is increasing there; if \( f'(x) < 0 \) , it decreases. As for our function, it is increasing on the interval \( (0, \frac{3}{2}) \) and decreasing on \( (\frac{3}{2}, \infty) \) . This simple but powerful analysis is crucial in finding where the function achieves its highest and lowest values on certain intervals.

Trigonometric Functions

Trigonometric functions (like sine and cosine in our example) oscillate and repeat over intervals, giving them unique properties in calculus. They are periodic, which means they have a specific interval after which the function starts to repeat its values.

The function in our exercise,

\( f(x) = \frac{1}{5}\text{sin}(x) + \frac{1}{4}\text{cos}(x) \)

combines sine and cosine, both periodic with a period of \( 2\pi \) radians. When dealing with trigonometric functions, their derivatives are also trigonometric:

• \( (\sin x)' = \frac{1}{5}\text{cos}(x) \)
• \( (\cos x)' = - \frac{1}{5}\text{sin}(x) \)

This interplay of sine and cosine is key to finding the critical points and understanding the function's behavior over an interval, such as \( (0, 2\pi) \) . The repetitive nature of such functions makes analyzing them interesting, especially since their high and low points (extrema) occur regularly, dictated by their period.