Problem 54 - 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. $$ f(x)=\frac{\sin x}{1+\cos ^{2} x} $$

Step-by-Step Solution

Answer

The function \( f(x)=\frac{\sin x}{1+\cos ^{2} x}\) is increasing on \((0, \pi)\) and decreasing on \((\pi, 2\pi)\), with a relative maximum at \(\pi\).

Step 1: Find derivative of the function

By using quotient rule (which states that the derivative of \((u/v)'= (vu' - uv')/v²\) where u,v are differentiable functions of x. Here, \(u=\sin x\) and \(v=1+\cos^{2} x\). Therefore \[f'(x)=\frac{(1+\cos^{2} x)\cos x -\sin x(-2\cos x \sin x)}{(1+\cos^{2} x)^{2}}\] which simplifies to \[f'(x)=\frac{\cos x +2\sin^{2}\cos x}{(1+\cos^{2} x)^{2}}\]

Step 2: Find the critical points

Setting \(f'(x)=0\) gives us \(x=0, \pi, 2\pi\). These are the potential points where the function could be increasing or decreasing.

Step 3: Apply First Derivative Test

Substitute values between the critical points in the derivative. Choose \(x= \pi/2, 3\pi/2\) for the intervals \((0, \pi)\) and \((\pi, 2\pi)\). The derivative is positive for \(x= \pi/2\) and negative for \(x=3\pi/2\). Hence, function is increasing on the interval \((0, \pi)\) and decreasing on \((\pi, 2\pi)\). Thus, there is a relative maximum at \((\pi, f(\pi))\).

Step 4: Confirm results graphically

By using a graphing utility, we can verify the findings above which should reflect that the function increases until \(\pi\) and then decreases onward, reaching a peak at \(\pi\)

Key Concepts

First Derivative Test

The First Derivative Test is an essential calculus concept used to determine where a function's relative extrema (the peaks and troughs) are located. An extremum can be a local maximum, where the function reaches a peak, or a local minimum, where it dips to its lowest point within a certain interval.

For the given function, after finding the derivative and setting it equal to zero, we discover potential extremal points. If the derivative changes from positive to negative at a point, it indicates a local maximum. Conversely, a change from negative to positive implies a local minimum. In the exercise, by evaluating the derivative at points between the critical values (like \(\pi/2\) and \(3\pi/2\)), we find that the sign of the first derivative changes from positive to negative at \(\pi\), suggesting a relative maximum there.

It's pivotal to remember that the First Derivative Test does not work at endpoints of intervals or where the derivative does not exist. However, in the interior of the interval where the derivative switches signs, the test is a reliable method to pinpoint extrema.

Increasing and Decreasing Functions

When exploring increasing and decreasing functions, calculus gives us a straightforward way to understand their behavior by looking at the first derivative. If the first derivative of a function, \(f'(x)\), is positive over an interval, it means the function is increasing on that interval. If \(f'(x)\) is negative, the function is decreasing.

In our exercise, we found that the first derivative is positive for \(x = \pi/2\) indicating that the function is increasing on the interval \(0, \pi\). And for \(x = 3\pi/2\) the derivative is negative, showing a decreasing trend on the interval \(\pi, 2\pi\). Understanding the nature of derivatives in relation to the original function is key to mastering calculus, as it directly tells us about the slope of the tangent line to the function's graph, and thus, about the function's increasing or decreasing tendency.

Critical Points

In calculus, critical points are the cornerstone for finding local maxima and minima; they are the x-values where the first derivative of a function equals zero or where the derivative does not exist. Identifying these points is essential because they are where the function could change direction.

In the context of the exercise, critical points were found by equating the derivative to zero and solving for \(x\). The solutions \(x = 0\), \(\pi\), and \(2\pi\) mark the critical points for our function. To determine what these points represent, we perform further tests, like the First Derivative Test, as previously discussed. It's important to note that not all critical points will yield extrema – some might be inflection points, where the function changes its concavity. This is just one of many nuances related to critical points in the study of calculus.